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Inverse Trigonometric Functions

Uttar Pradesh Board · Class 12 · Mathematics

NCERT Solutions for Inverse Trigonometric Functions — Uttar Pradesh Board Class 12 Mathematics.

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43 Questions Solved · 3 Sections

Exercise 2.1

1Find the principal value of sin⁡−1(−12)\sin^{-1}\left(-\dfrac{1}{2}\right).Show solution
Given: sin⁡−1(−12)\sin^{-1}\left(-\dfrac{1}{2}\right)

Concept: The principal value branch of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right].

Working:

Let sin⁡−1(−12)=y\sin^{-1}\left(-\dfrac{1}{2}\right) = y. Then sin⁡y=−12\sin y = -\dfrac{1}{2}.

We know that sin⁡(−π6)=−12\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2} and −π6∈[−π2,π2]-\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right].

Answer: The principal value of sin⁡−1(−12)=−π6\sin^{-1}\left(-\dfrac{1}{2}\right) = \boxed{-\dfrac{\pi}{6}}.
2Find the principal value of cos⁡−1(32)\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right).Show solution
Given: cos⁡−1(32)\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)

Concept: The principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi].

Working:

Let cos⁡−1(32)=y\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = y. Then cos⁡y=32\cos y = \dfrac{\sqrt{3}}{2}.

We know that cos⁡(π6)=32\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2} and π6∈[0,π]\dfrac{\pi}{6} \in [0, \pi].

Answer: The principal value of cos⁡−1(32)=π6\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = \boxed{\dfrac{\pi}{6}}.
3Find the principal value of csc⁡−1(2)\csc^{-1}(2).Show solution
Given: csc⁡−1(2)\csc^{-1}(2)

Concept: The principal value branch of csc⁡−1\csc^{-1} is [−π2,π2]−{0}\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}.

Working:

Let csc⁡−1(2)=y\csc^{-1}(2) = y. Then csc⁡y=2\csc y = 2, i.e., sin⁡y=12\sin y = \dfrac{1}{2}.

We know that sin⁡(π6)=12\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}, so csc⁡(π6)=2\csc\left(\dfrac{\pi}{6}\right) = 2 and π6∈[−π2,π2]−{0}\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}.

Answer: The principal value of csc⁡−1(2)=π6\csc^{-1}(2) = \boxed{\dfrac{\pi}{6}}.
4Find the principal value of tan⁡−1(−3)\tan^{-1}(-\sqrt{3}).Show solution
Given: tan⁡−1(−3)\tan^{-1}(-\sqrt{3})

Concept: The principal value branch of tan⁡−1\tan^{-1} is (−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Working:

Let tan⁡−1(−3)=y\tan^{-1}(-\sqrt{3}) = y. Then tan⁡y=−3\tan y = -\sqrt{3}.

We know that tan⁡(−π3)=−3\tan\left(-\dfrac{\pi}{3}\right) = -\sqrt{3} and −π3∈(−π2,π2)-\dfrac{\pi}{3} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Answer: The principal value of tan⁡−1(−3)=−π3\tan^{-1}(-\sqrt{3}) = \boxed{-\dfrac{\pi}{3}}.
5Find the principal value of cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{2}\right).Show solution
Given: cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{2}\right)

Concept: The principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi].

Working:

Let cos⁡−1(−12)=y\cos^{-1}\left(-\dfrac{1}{2}\right) = y. Then cos⁡y=−12\cos y = -\dfrac{1}{2}.

We know that cos⁡(2π3)=−12\cos\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{2} and 2π3∈[0,π]\dfrac{2\pi}{3} \in [0, \pi].

Answer: The principal value of cos⁡−1(−12)=2π3\cos^{-1}\left(-\dfrac{1}{2}\right) = \boxed{\dfrac{2\pi}{3}}.
6Find the principal value of tan⁡−1(−1)\tan^{-1}(-1).Show solution
Given: tan⁡−1(−1)\tan^{-1}(-1)

Concept: The principal value branch of tan⁡−1\tan^{-1} is (−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Working:

Let tan⁡−1(−1)=y\tan^{-1}(-1) = y. Then tan⁡y=−1\tan y = -1.

We know that tan⁡(−π4)=−1\tan\left(-\dfrac{\pi}{4}\right) = -1 and −π4∈(−π2,π2)-\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Answer: The principal value of tan⁡−1(−1)=−π4\tan^{-1}(-1) = \boxed{-\dfrac{\pi}{4}}.
7Find the principal value of sec⁡−1(23)\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right).Show solution
Given: sec⁡−1(23)\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right)

Concept: The principal value branch of sec⁡−1\sec^{-1} is [0,π]−{π2}[0, \pi] - \left\{\dfrac{\pi}{2}\right\}.

Working:

Let sec⁡−1(23)=y\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = y. Then sec⁡y=23\sec y = \dfrac{2}{\sqrt{3}}, i.e., cos⁡y=32\cos y = \dfrac{\sqrt{3}}{2}.

We know that cos⁡(π6)=32\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2} and π6∈[0,π]−{π2}\dfrac{\pi}{6} \in [0, \pi] - \left\{\dfrac{\pi}{2}\right\}.

Answer: The principal value of sec⁡−1(23)=π6\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = \boxed{\dfrac{\pi}{6}}.
8Find the principal value of cot⁡−1(3)\cot^{-1}(\sqrt{3}).Show solution
Given: cot⁡−1(3)\cot^{-1}(\sqrt{3})

Concept: The principal value branch of cot⁡−1\cot^{-1} is (0,π)(0, \pi).

Working:

Let cot⁡−1(3)=y\cot^{-1}(\sqrt{3}) = y. Then cot⁡y=3\cot y = \sqrt{3}.

We know that cot⁡(π6)=3\cot\left(\dfrac{\pi}{6}\right) = \sqrt{3} and π6∈(0,π)\dfrac{\pi}{6} \in (0, \pi).

Answer: The principal value of cot⁡−1(3)=π6\cot^{-1}(\sqrt{3}) = \boxed{\dfrac{\pi}{6}}.
9Find the principal value of cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right).Show solution
Given: cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)

Concept: The principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi].

Working:

Let cos⁡−1(−12)=y\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = y. Then cos⁡y=−12\cos y = -\dfrac{1}{\sqrt{2}}.

We know that cos⁡(3π4)=−12\cos\left(\dfrac{3\pi}{4}\right) = -\dfrac{1}{\sqrt{2}} and 3π4∈[0,π]\dfrac{3\pi}{4} \in [0, \pi].

Answer: The principal value of cos⁡−1(−12)=3π4\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = \boxed{\dfrac{3\pi}{4}}.
10Find the principal value of csc⁡−1(−2)\csc^{-1}(-\sqrt{2}).Show solution
Given: csc⁡−1(−2)\csc^{-1}(-\sqrt{2})

Concept: The principal value branch of csc⁡−1\csc^{-1} is [−π2,π2]−{0}\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}.

Working:

Let csc⁡−1(−2)=y\csc^{-1}(-\sqrt{2}) = y. Then csc⁡y=−2\csc y = -\sqrt{2}, i.e., sin⁡y=−12\sin y = -\dfrac{1}{\sqrt{2}}.

We know that sin⁡(−π4)=−12\sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}} and −π4∈[−π2,π2]−{0}-\dfrac{\pi}{4} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}.

Answer: The principal value of csc⁡−1(−2)=−π4\csc^{-1}(-\sqrt{2}) = \boxed{-\dfrac{\pi}{4}}.
11Find the value of tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right).Show solution
Given: tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right)

Working:

Step 1: Find tan⁡−1(1)\tan^{-1}(1).
tan⁡(π4)=1⇒tan⁡−1(1)=π4\tan\left(\frac{\pi}{4}\right) = 1 \Rightarrow \tan^{-1}(1) = \frac{\pi}{4}

Step 2: Find cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{2}\right).
cos⁡(2π3)=−12⇒cos⁡−1(−12)=2π3\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \Rightarrow \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}

Step 3: Find sin⁡−1(−12)\sin^{-1}\left(-\dfrac{1}{2}\right).
sin⁡(−π6)=−12⇒sin⁡−1(−12)=−π6\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \Rightarrow \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}

Step 4: Add all three values.
π4+2π3+(−π6)=3π12+8π12−2π12=9π12=3π4\frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right) = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}

Answer: tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)=3π4\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right) = \boxed{\dfrac{3\pi}{4}}.
12Find the value of cos⁡−112+2sin⁡−112\cos^{-1}\dfrac{1}{2} + 2\sin^{-1}\dfrac{1}{2}.Show solution
Given: cos⁡−112+2sin⁡−112\cos^{-1}\dfrac{1}{2} + 2\sin^{-1}\dfrac{1}{2}

Working:

Step 1: Find cos⁡−112\cos^{-1}\dfrac{1}{2}.
cos⁡(π3)=12⇒cos⁡−112=π3\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \Rightarrow \cos^{-1}\frac{1}{2} = \frac{\pi}{3}

Step 2: Find sin⁡−112\sin^{-1}\dfrac{1}{2}.
sin⁡(π6)=12⇒sin⁡−112=π6\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \Rightarrow \sin^{-1}\frac{1}{2} = \frac{\pi}{6}

Step 3: Compute the expression.
π3+2×π6=π3+π3=2π3\frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}

Answer: cos⁡−112+2sin⁡−112=2π3\cos^{-1}\dfrac{1}{2} + 2\sin^{-1}\dfrac{1}{2} = \boxed{\dfrac{2\pi}{3}}.
13If sin⁡−1x=y\sin^{-1}x = y, then
(A) 0≤y≤π0 \leq y \leq \pi
(B) −π2≤y≤π2-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}
(C) 0 < y < \pi
(D) -\dfrac{\pi}{2} < y < \dfrac{\pi}{2}Show solution
Correct Option: (B) −π2≤y≤π2-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}

Justification: The principal value branch of sin⁡−1\sin^{-1} is defined as [−π2,π2]\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right], which is a closed interval. Therefore, if sin⁡−1x=y\sin^{-1}x = y, then y∈[−π2,π2]y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right], i.e., −π2≤y≤π2-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}.
14tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt{3} - \sec^{-1}(-2) is equal to
(A) π\pi
(B) −π3-\dfrac{\pi}{3}
(C) π3\dfrac{\pi}{3}
(D) 2π3\dfrac{2\pi}{3}Show solution
Correct Option: (B) −π3-\dfrac{\pi}{3}

Working:

Step 1: Find tan⁡−13\tan^{-1}\sqrt{3}.
tan⁡(π3)=3⇒tan⁡−13=π3\tan\left(\frac{\pi}{3}\right) = \sqrt{3} \Rightarrow \tan^{-1}\sqrt{3} = \frac{\pi}{3}

Step 2: Find sec⁡−1(−2)\sec^{-1}(-2).
sec⁡y=−2⇒cos⁡y=−12⇒y=2π3∈[0,π]−{π2}\sec y = -2 \Rightarrow \cos y = -\frac{1}{2} \Rightarrow y = \frac{2\pi}{3} \in [0,\pi]-\left\{\frac{\pi}{2}\right\}
∴sec⁡−1(−2)=2π3\therefore \sec^{-1}(-2) = \frac{2\pi}{3}

Step 3: Compute.
tan⁡−13−sec⁡−1(−2)=π3−2π3=−π3\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}

Answer: Option (B) −π3-\dfrac{\pi}{3}.

Exercise 2.2

1Prove that 3sin⁡−1x=sin⁡−1(3x−4x3)3\sin^{-1}x = \sin^{-1}(3x - 4x^3), x∈[−12,12]x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right].Show solution
To Prove: 3sin⁡−1x=sin⁡−1(3x−4x3)3\sin^{-1}x = \sin^{-1}(3x - 4x^3)

Proof:

Let sin⁡−1x=θ\sin^{-1}x = \theta, so x=sin⁡θx = \sin\theta.

Since x∈[−12,12]x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right], we have θ∈[−π6,π6]\theta \in \left[-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right], which means 3θ∈[−π2,π2]3\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right].

Now, using the triple angle formula:
sin⁡3θ=3sin⁡θ−4sin⁡3θ=3x−4x3\sin 3\theta = 3\sin\theta - 4\sin^3\theta = 3x - 4x^3

Since 3θ∈[−π2,π2]3\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right], we can apply sin⁡−1\sin^{-1} to both sides:
3θ=sin⁡−1(3x−4x3)3\theta = \sin^{-1}(3x - 4x^3)

Substituting back θ=sin⁡−1x\theta = \sin^{-1}x:
3sin⁡−1x=sin⁡−1(3x−4x3)Hence Proved.3\sin^{-1}x = \sin^{-1}(3x - 4x^3) \qquad \textbf{Hence Proved.}
2Prove that 3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x = \cos^{-1}(4x^3 - 3x), x∈[12,1]x \in \left[\dfrac{1}{2}, 1\right].Show solution
To Prove: 3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)

Proof:

Let cos⁡−1x=θ\cos^{-1}x = \theta, so x=cos⁡θx = \cos\theta.

Since x∈[12,1]x \in \left[\dfrac{1}{2}, 1\right], we have θ∈[0,π3]\theta \in \left[0, \dfrac{\pi}{3}\right], which means 3θ∈[0,π]3\theta \in [0, \pi].

Using the triple angle formula:
cos⁡3θ=4cos⁡3θ−3cos⁡θ=4x3−3x\cos 3\theta = 4\cos^3\theta - 3\cos\theta = 4x^3 - 3x

Since 3θ∈[0,π]3\theta \in [0, \pi], we can apply cos⁡−1\cos^{-1} to both sides:
3θ=cos⁡−1(4x3−3x)3\theta = \cos^{-1}(4x^3 - 3x)

Substituting back θ=cos⁡−1x\theta = \cos^{-1}x:
3cos⁡−1x=cos⁡−1(4x3−3x)Hence Proved.3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \qquad \textbf{Hence Proved.}
3Write tan⁡−11+x2−1x\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}, x≠0x \neq 0 in the simplest form.Show solution
Given: tan⁡−11+x2−1x\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}

Working:

Let x=tan⁡θx = \tan\theta, so θ=tan⁡−1x\theta = \tan^{-1}x, where θ∈(−π2,π2)\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Then:
1+x2=1+tan⁡2θ=sec⁡θ\sqrt{1+x^2} = \sqrt{1+\tan^2\theta} = \sec\theta

Substituting:
tan⁡−1sec⁡θ−1tan⁡θ=tan⁡−11−cos⁡θsin⁡θ\tan^{-1}\frac{\sec\theta - 1}{\tan\theta} = \tan^{-1}\frac{1 - \cos\theta}{\sin\theta}

Using half-angle identities: 1−cos⁡θ=2sin⁡2θ21 - \cos\theta = 2\sin^2\dfrac{\theta}{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}:
=tan⁡−12sin⁡2θ22sin⁡θ2cos⁡θ2=tan⁡−1(tan⁡θ2)=θ2= \tan^{-1}\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2}

Substituting back θ=tan⁡−1x\theta = \tan^{-1}x:
=12tan⁡−1x= \frac{1}{2}\tan^{-1}x

Answer: tan⁡−11+x2−1x=12tan⁡−1x\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x} = \dfrac{1}{2}\tan^{-1}x.
4Write tan⁡−1(1−cos⁡x1+cos⁡x)\tan^{-1}\left(\sqrt{\dfrac{1-\cos x}{1+\cos x}}\right), 0 < x < \pi in the simplest form.Show solution
Given: tan⁡−1(1−cos⁡x1+cos⁡x)\tan^{-1}\left(\sqrt{\dfrac{1-\cos x}{1+\cos x}}\right), 0 < x < \pi

Working:

Using half-angle identities:
1−cos⁡x=2sin⁡2x2,1+cos⁡x=2cos⁡2x21 - \cos x = 2\sin^2\frac{x}{2}, \quad 1 + \cos x = 2\cos^2\frac{x}{2}

So:
1−cos⁡x1+cos⁡x=2sin⁡2x22cos⁡2x2=∣tan⁡x2∣\sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} = \left|\tan\frac{x}{2}\right|

Since 0 < x < \pi, we have 0 < \dfrac{x}{2} < \dfrac{\pi}{2}, so \tan\dfrac{x}{2} > 0.

Therefore:
tan⁡−1(tan⁡x2)=x2\tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}

Answer: tan⁡−1(1−cos⁡x1+cos⁡x)=x2\tan^{-1}\left(\sqrt{\dfrac{1-\cos x}{1+\cos x}}\right) = \dfrac{x}{2}.
5Write tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right), -\dfrac{\pi}{4} < x < \dfrac{3\pi}{4} in the simplest form.Show solution
Given: tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)

Working:

Divide numerator and denominator by cos⁡x\cos x:
tan⁡−1(1−tan⁡x1+tan⁡x)\tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right)

Using the identity tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B} with A=π4A = \dfrac{\pi}{4} and B=xB = x:
=tan⁡−1(tan⁡(π4−x))= \tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right)

Since -\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}, we have -\pi < \dfrac{\pi}{4} - x < \dfrac{\pi}{2}, but more precisely π4−x∈(−π2,π2)\dfrac{\pi}{4} - x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) for the given range.

Therefore:
=π4−x= \frac{\pi}{4} - x

Answer: tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)=π4−x\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right) = \dfrac{\pi}{4} - x.
6Write tan⁡−1xa2−x2\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}}, |x| < a in the simplest form.Show solution
Given: tan⁡−1xa2−x2\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}}, |x| < a

Working:

Let x=asin⁡θx = a\sin\theta, so θ=sin⁡−1xa\theta = \sin^{-1}\dfrac{x}{a}, where θ∈(−π2,π2)\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Then:
a2−x2=a2−a2sin⁡2θ=acos⁡θ\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta

Substituting:
tan⁡−1asin⁡θacos⁡θ=tan⁡−1(tan⁡θ)=θ=sin⁡−1xa\tan^{-1}\frac{a\sin\theta}{a\cos\theta} = \tan^{-1}(\tan\theta) = \theta = \sin^{-1}\frac{x}{a}

Answer: tan⁡−1xa2−x2=sin⁡−1xa\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}.
7Write tan⁡−1(3a2x−x3a3−3ax2)\tan^{-1}\left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right), a > 0; -\dfrac{a}{\sqrt{3}} < x < \dfrac{a}{\sqrt{3}} in the simplest form.Show solution
Given: tan⁡−1(3a2x−x3a3−3ax2)\tan^{-1}\left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right)

Working:

Let x=atan⁡θx = a\tan\theta, so θ=tan⁡−1xa\theta = \tan^{-1}\dfrac{x}{a}.

Since -\dfrac{a}{\sqrt{3}} < x < \dfrac{a}{\sqrt{3}}, we have -\dfrac{1}{\sqrt{3}} < \tan\theta < \dfrac{1}{\sqrt{3}}, so θ∈(−π6,π6)\theta \in \left(-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right) and 3θ∈(−π2,π2)3\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right).

Substituting x=atan⁡θx = a\tan\theta:
3a2(atan⁡θ)−(atan⁡θ)3a3−3a(atan⁡θ)2=3a3tan⁡θ−a3tan⁡3θa3−3a3tan⁡2θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ=tan⁡3θ\frac{3a^2(a\tan\theta) - (a\tan\theta)^3}{a^3 - 3a(a\tan\theta)^2} = \frac{3a^3\tan\theta - a^3\tan^3\theta}{a^3 - 3a^3\tan^2\theta} = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta

Therefore:
tan⁡−1(tan⁡3θ)=3θ=3tan⁡−1xa\tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}\frac{x}{a}

Answer: tan⁡−1(3a2x−x3a3−3ax2)=3tan⁡−1xa\tan^{-1}\left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right) = 3\tan^{-1}\dfrac{x}{a}.
8Find the value of tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac{1}{2}\right)\right].Show solution
Given: tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac{1}{2}\right)\right]

Working:

Step 1: Find sin⁡−112\sin^{-1}\dfrac{1}{2}.
sin⁡−112=π6\sin^{-1}\frac{1}{2} = \frac{\pi}{6}

Step 2: Find 2sin⁡−1122\sin^{-1}\dfrac{1}{2}.
2×π6=π32 \times \frac{\pi}{6} = \frac{\pi}{3}

Step 3: Find cos⁡(π3)\cos\left(\dfrac{\pi}{3}\right).
cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}

Step 4: Compute the full expression.
tan⁡−1[2×12]=tan⁡−1(1)=π4\tan^{-1}\left[2 \times \frac{1}{2}\right] = \tan^{-1}(1) = \frac{\pi}{4}

Answer: tan⁡−1[2cos⁡(2sin⁡−112)]=π4\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac{1}{2}\right)\right] = \boxed{\dfrac{\pi}{4}}.
9Find the value of tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2]\tan\dfrac{1}{2}\left[\sin^{-1}\dfrac{2x}{1+x^2} + \cos^{-1}\dfrac{1-y^2}{1+y^2}\right], |x| < 1, y > 0 and xy < 1.Show solution
Given: tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2]\tan\dfrac{1}{2}\left[\sin^{-1}\dfrac{2x}{1+x^2} + \cos^{-1}\dfrac{1-y^2}{1+y^2}\right]

Working:

Step 1: Simplify sin⁡−12x1+x2\sin^{-1}\dfrac{2x}{1+x^2}.

Let x=tan⁡αx = \tan\alpha, so α=tan⁡−1x\alpha = \tan^{-1}x. Since |x| < 1, α∈(−π4,π4)\alpha \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right).
sin⁡−12tan⁡α1+tan⁡2α=sin⁡−1(sin⁡2α)=2α=2tan⁡−1x\sin^{-1}\frac{2\tan\alpha}{1+\tan^2\alpha} = \sin^{-1}(\sin 2\alpha) = 2\alpha = 2\tan^{-1}x

Step 2: Simplify cos⁡−11−y21+y2\cos^{-1}\dfrac{1-y^2}{1+y^2}.

Let y=tan⁡βy = \tan\beta, so β=tan⁡−1y\beta = \tan^{-1}y. Since y > 0, β∈(0,π2)\beta \in \left(0, \dfrac{\pi}{2}\right).
cos⁡−11−tan⁡2β1+tan⁡2β=cos⁡−1(cos⁡2β)=2β=2tan⁡−1y\cos^{-1}\frac{1-\tan^2\beta}{1+\tan^2\beta} = \cos^{-1}(\cos 2\beta) = 2\beta = 2\tan^{-1}y

Step 3: Substitute back.
tan⁡12[2tan⁡−1x+2tan⁡−1y]=tan⁡[tan⁡−1x+tan⁡−1y]\tan\frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y\right] = \tan\left[\tan^{-1}x + \tan^{-1}y\right]

Step 4: Use the addition formula for tan⁡−1\tan^{-1}. Since xy < 1:
tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}

Therefore:
tan⁡(tan⁡−1x+y1−xy)=x+y1−xy\tan\left(\tan^{-1}\frac{x+y}{1-xy}\right) = \frac{x+y}{1-xy}

Answer: tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2]=x+y1−xy\tan\dfrac{1}{2}\left[\sin^{-1}\dfrac{2x}{1+x^2} + \cos^{-1}\dfrac{1-y^2}{1+y^2}\right] = \dfrac{x+y}{1-xy}.
10Find the value of sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right).Show solution
Given: sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right)

Working:

Note that 2π3∉[−π2,π2]\dfrac{2\pi}{3} \notin \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right], so we cannot directly apply sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x.

We write:
sin⁡2π3=sin⁡(π−2π3)=sin⁡π3\sin\frac{2\pi}{3} = \sin\left(\pi - \frac{2\pi}{3}\right) = \sin\frac{\pi}{3}

Since π3∈[−π2,π2]\dfrac{\pi}{3} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]:
sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡π3)=π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3}

Answer: sin⁡−1(sin⁡2π3)=π3\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) = \boxed{\dfrac{\pi}{3}}.
11Find the value of tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right).Show solution
Given: tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)

Working:

Note that 3π4∉(−π2,π2)\dfrac{3\pi}{4} \notin \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), so we cannot directly apply tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x.

We write:
tan⁡3π4=tan⁡(π−π4)=−tan⁡π4=−1\tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\frac{\pi}{4} = -1

Since −π4∈(−π2,π2)-\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right):
tan⁡−1(tan⁡3π4)=tan⁡−1(−1)=−π4\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}

Answer: tan⁡−1(tan⁡3π4)=−π4\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right) = \boxed{-\dfrac{\pi}{4}}.
12Find the value of tan⁡(sin⁡−135+cot⁡−132)\tan\left(\sin^{-1}\dfrac{3}{5} + \cot^{-1}\dfrac{3}{2}\right).Show solution
Given: tan⁡(sin⁡−135+cot⁡−132)\tan\left(\sin^{-1}\dfrac{3}{5} + \cot^{-1}\dfrac{3}{2}\right)

Working:

Step 1: Let sin⁡−135=α\sin^{-1}\dfrac{3}{5} = \alpha, so sin⁡α=35\sin\alpha = \dfrac{3}{5}.

Using Pythagoras: cos⁡α=45\cos\alpha = \dfrac{4}{5}, so tan⁡α=34\tan\alpha = \dfrac{3}{4}.

Step 2: Let cot⁡−132=β\cot^{-1}\dfrac{3}{2} = \beta, so cot⁡β=32\cot\beta = \dfrac{3}{2}, which gives tan⁡β=23\tan\beta = \dfrac{2}{3}.

Step 3: Use the addition formula:
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=34+231−34⋅23\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\dfrac{3}{4} + \dfrac{2}{3}}{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}

=912+8121−612=1712612=176= \frac{\dfrac{9}{12} + \dfrac{8}{12}}{1 - \dfrac{6}{12}} = \frac{\dfrac{17}{12}}{\dfrac{6}{12}} = \frac{17}{6}

Answer: tan⁡(sin⁡−135+cot⁡−132)=176\tan\left(\sin^{-1}\dfrac{3}{5} + \cot^{-1}\dfrac{3}{2}\right) = \boxed{\dfrac{17}{6}}.
13cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right) is equal to
(A) 7π6\dfrac{7\pi}{6}
(B) 5π6\dfrac{5\pi}{6}
(C) π3\dfrac{\pi}{3}
(D) π6\dfrac{\pi}{6}Show solution
Correct Option: (B) 5π6\dfrac{5\pi}{6}

Working:

7π6∉[0,π]\dfrac{7\pi}{6} \notin [0, \pi], so we cannot directly apply cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x.

cos⁡7π6=cos⁡(2π−7π6)=cos⁡5π6\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}

Alternatively: cos⁡7π6=cos⁡(π+π6)=−cos⁡π6=−32\cos\dfrac{7\pi}{6} = \cos\left(\pi + \dfrac{\pi}{6}\right) = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}

And cos⁡5π6=cos⁡(π−π6)=−cos⁡π6=−32\cos\dfrac{5\pi}{6} = \cos\left(\pi - \dfrac{\pi}{6}\right) = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}

Since 5π6∈[0,π]\dfrac{5\pi}{6} \in [0, \pi]:
cos⁡−1(cos⁡7π6)=5π6\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}

Answer: Option (B) 5π6\dfrac{5\pi}{6}.
14sin⁡(π3−sin⁡−1(−12))\sin\left(\dfrac{\pi}{3} - \sin^{-1}\left(-\dfrac{1}{2}\right)\right) is equal to
(A) 12\dfrac{1}{2}
(B) 13\dfrac{1}{3}
(C) 14\dfrac{1}{4}
(D) 11Show solution
Correct Option: (D) 11

Working:

Step 1: Find sin⁡−1(−12)\sin^{-1}\left(-\dfrac{1}{2}\right).
sin⁡−1(−12)=−π6\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}

Step 2: Substitute.
sin⁡(π3−(−π6))=sin⁡(π3+π6)=sin⁡(2π6+π6)=sin⁡3π6=sin⁡π2=1\sin\left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right) = \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \sin\left(\frac{2\pi}{6} + \frac{\pi}{6}\right) = \sin\frac{3\pi}{6} = \sin\frac{\pi}{2} = 1

Answer: Option (D) 11.
15tan⁡−13−cot⁡−1(−3)\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) is equal to
(A) π\pi
(B) −π2-\dfrac{\pi}{2}
(C) 00
(D) 232\sqrt{3}Show solution
Correct Option: (B) −π2-\dfrac{\pi}{2}

Working:

Step 1: Find tan⁡−13\tan^{-1}\sqrt{3}.
tan⁡−13=π3\tan^{-1}\sqrt{3} = \frac{\pi}{3}

Step 2: Find cot⁡−1(−3)\cot^{-1}(-\sqrt{3}).
cot⁡y=−3=−cot⁡π6=cot⁡(π−π6)=cot⁡5π6\cot y = -\sqrt{3} = -\cot\frac{\pi}{6} = \cot\left(\pi - \frac{\pi}{6}\right) = \cot\frac{5\pi}{6}
Since 5π6∈(0,π)\dfrac{5\pi}{6} \in (0, \pi): cot⁡−1(−3)=5π6\cot^{-1}(-\sqrt{3}) = \dfrac{5\pi}{6}

Step 3: Compute.
tan⁡−13−cot⁡−1(−3)=π3−5π6=2π6−5π6=−3π6=−π2\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6} = \frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}

Answer: Option (B) −π2-\dfrac{\pi}{2}.

Miscellaneous Exercise on Chapter 2

1Find the value of cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right).Show solution
Given: cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)

Working:

13π6∉[0,π]\dfrac{13\pi}{6} \notin [0, \pi], so we simplify:
cos⁡13π6=cos⁡(2π+π6)=cos⁡π6\cos\frac{13\pi}{6} = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\frac{\pi}{6}

Since π6∈[0,π]\dfrac{\pi}{6} \in [0, \pi]:
cos⁡−1(cos⁡13π6)=cos⁡−1(cos⁡π6)=π6\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\left(\cos\frac{\pi}{6}\right) = \frac{\pi}{6}

Answer: cos⁡−1(cos⁡13π6)=π6\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right) = \boxed{\dfrac{\pi}{6}}.
2Find the value of tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right).Show solution
Given: tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right)

Working:

7π6∉(−π2,π2)\dfrac{7\pi}{6} \notin \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), so we simplify:
tan⁡7π6=tan⁡(π+π6)=tan⁡π6\tan\frac{7\pi}{6} = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\frac{\pi}{6}

Since π6∈(−π2,π2)\dfrac{\pi}{6} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right):
tan⁡−1(tan⁡7π6)=tan⁡−1(tan⁡π6)=π6\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\left(\tan\frac{\pi}{6}\right) = \frac{\pi}{6}

Answer: tan⁡−1(tan⁡7π6)=π6\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right) = \boxed{\dfrac{\pi}{6}}.
3Prove that 2sin⁡−135=tan⁡−12472\sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{24}{7}.Show solution
To Prove: 2sin⁡−135=tan⁡−12472\sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{24}{7}

Proof:

Let sin⁡−135=θ\sin^{-1}\dfrac{3}{5} = \theta, so sin⁡θ=35\sin\theta = \dfrac{3}{5}.

Using Pythagoras: cos⁡θ=45\cos\theta = \dfrac{4}{5}, so tan⁡θ=34\tan\theta = \dfrac{3}{4}.

Now:
tan⁡2θ=2tan⁡θ1−tan⁡2θ=2⋅341−(34)2=321−916=32716=32×167=247\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2 \cdot \dfrac{3}{4}}{1 - \left(\dfrac{3}{4}\right)^2} = \frac{\dfrac{3}{2}}{1 - \dfrac{9}{16}} = \frac{\dfrac{3}{2}}{\dfrac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}

Since θ∈(0,π2)\theta \in \left(0, \dfrac{\pi}{2}\right), we have 2θ∈(0,π)2\theta \in (0, \pi). Also \tan 2\theta = \dfrac{24}{7} > 0, so 2θ∈(0,π2)2\theta \in \left(0, \dfrac{\pi}{2}\right).

Therefore: 2θ=tan⁡−12472\theta = \tan^{-1}\dfrac{24}{7}, i.e., 2sin⁡−135=tan⁡−12472\sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{24}{7}. Hence Proved.\qquad \textbf{Hence Proved.}
4Prove that sin⁡−1817+sin⁡−135=tan⁡−17736\sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{77}{36}.Show solution
To Prove: sin⁡−1817+sin⁡−135=tan⁡−17736\sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{77}{36}

Proof:

Let sin⁡−1817=α\sin^{-1}\dfrac{8}{17} = \alpha and sin⁡−135=β\sin^{-1}\dfrac{3}{5} = \beta.

So sin⁡α=817\sin\alpha = \dfrac{8}{17}, cos⁡α=1517\cos\alpha = \dfrac{15}{17}, tan⁡α=815\tan\alpha = \dfrac{8}{15}.

And sin⁡β=35\sin\beta = \dfrac{3}{5}, cos⁡β=45\cos\beta = \dfrac{4}{5}, tan⁡β=34\tan\beta = \dfrac{3}{4}.

Now:
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=815+341−815⋅34=3260+45601−2460=77603660=7736\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\dfrac{8}{15} + \dfrac{3}{4}}{1 - \dfrac{8}{15} \cdot \dfrac{3}{4}} = \frac{\dfrac{32}{60} + \dfrac{45}{60}}{1 - \dfrac{24}{60}} = \frac{\dfrac{77}{60}}{\dfrac{36}{60}} = \frac{77}{36}

Since α,β∈(0,π2)\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right) and \tan(\alpha+\beta) > 0, we have α+β∈(0,π2)\alpha + \beta \in \left(0, \dfrac{\pi}{2}\right).

Therefore: α+β=tan⁡−17736\alpha + \beta = \tan^{-1}\dfrac{77}{36}, i.e., sin⁡−1817+sin⁡−135=tan⁡−17736\sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{77}{36}. Hence Proved.\qquad \textbf{Hence Proved.}
5Prove that cos⁡−145+cos⁡−11213=cos⁡−13365\cos^{-1}\dfrac{4}{5} + \cos^{-1}\dfrac{12}{13} = \cos^{-1}\dfrac{33}{65}.Show solution
To Prove: cos⁡−145+cos⁡−11213=cos⁡−13365\cos^{-1}\dfrac{4}{5} + \cos^{-1}\dfrac{12}{13} = \cos^{-1}\dfrac{33}{65}

Proof:

Let cos⁡−145=α\cos^{-1}\dfrac{4}{5} = \alpha and cos⁡−11213=β\cos^{-1}\dfrac{12}{13} = \beta.

So cos⁡α=45\cos\alpha = \dfrac{4}{5}, sin⁡α=35\sin\alpha = \dfrac{3}{5}.

And cos⁡β=1213\cos\beta = \dfrac{12}{13}, sin⁡β=513\sin\beta = \dfrac{5}{13}.

Using the cosine addition formula:
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β=45⋅1213−35⋅513\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13}
=4865−1565=3365= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}

Since α,β∈(0,π2)\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right), we have α+β∈(0,π)\alpha + \beta \in (0, \pi), which is the range of cos⁡−1\cos^{-1}.

Therefore: α+β=cos⁡−13365\alpha + \beta = \cos^{-1}\dfrac{33}{65}, i.e., cos⁡−145+cos⁡−11213=cos⁡−13365\cos^{-1}\dfrac{4}{5} + \cos^{-1}\dfrac{12}{13} = \cos^{-1}\dfrac{33}{65}. Hence Proved.\qquad \textbf{Hence Proved.}
6Prove that cos⁡−11213+sin⁡−135=sin⁡−15665\cos^{-1}\dfrac{12}{13} + \sin^{-1}\dfrac{3}{5} = \sin^{-1}\dfrac{56}{65}.Show solution
To Prove: cos⁡−11213+sin⁡−135=sin⁡−15665\cos^{-1}\dfrac{12}{13} + \sin^{-1}\dfrac{3}{5} = \sin^{-1}\dfrac{56}{65}

Proof:

Let cos⁡−11213=α\cos^{-1}\dfrac{12}{13} = \alpha and sin⁡−135=β\sin^{-1}\dfrac{3}{5} = \beta.

So cos⁡α=1213\cos\alpha = \dfrac{12}{13}, sin⁡α=513\sin\alpha = \dfrac{5}{13}.

And sin⁡β=35\sin\beta = \dfrac{3}{5}, cos⁡β=45\cos\beta = \dfrac{4}{5}.

Using the sine addition formula:
sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β=513⋅45+1213⋅35\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5}
=2065+3665=5665= \frac{20}{65} + \frac{36}{65} = \frac{56}{65}

Since α,β∈(0,π2)\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right), we have α+β∈(0,π)\alpha + \beta \in \left(0, \pi\right). Also \sin(\alpha+\beta) > 0, so α+β∈(0,π2)\alpha + \beta \in \left(0, \dfrac{\pi}{2}\right), which lies in the range of sin⁡−1\sin^{-1}.

Therefore: α+β=sin⁡−15665\alpha + \beta = \sin^{-1}\dfrac{56}{65}, i.e., cos⁡−11213+sin⁡−135=sin⁡−15665\cos^{-1}\dfrac{12}{13} + \sin^{-1}\dfrac{3}{5} = \sin^{-1}\dfrac{56}{65}. Hence Proved.\qquad \textbf{Hence Proved.}
7Prove that tan⁡−16316=sin⁡−1513+cos⁡−135\tan^{-1}\dfrac{63}{16} = \sin^{-1}\dfrac{5}{13} + \cos^{-1}\dfrac{3}{5}.Show solution
To Prove: tan⁡−16316=sin⁡−1513+cos⁡−135\tan^{-1}\dfrac{63}{16} = \sin^{-1}\dfrac{5}{13} + \cos^{-1}\dfrac{3}{5}

Proof:

Let sin⁡−1513=α\sin^{-1}\dfrac{5}{13} = \alpha and cos⁡−135=β\cos^{-1}\dfrac{3}{5} = \beta.

So sin⁡α=513\sin\alpha = \dfrac{5}{13}, cos⁡α=1213\cos\alpha = \dfrac{12}{13}, tan⁡α=512\tan\alpha = \dfrac{5}{12}.

And cos⁡β=35\cos\beta = \dfrac{3}{5}, sin⁡β=45\sin\beta = \dfrac{4}{5}, tan⁡β=43\tan\beta = \dfrac{4}{3}.

Using the tangent addition formula:
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=512+431−512⋅43=512+16121−2036=21121636=211249=2112×94=18948=6316\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\dfrac{5}{12} + \dfrac{4}{3}}{1 - \dfrac{5}{12} \cdot \dfrac{4}{3}} = \frac{\dfrac{5}{12} + \dfrac{16}{12}}{1 - \dfrac{20}{36}} = \frac{\dfrac{21}{12}}{\dfrac{16}{36}} = \frac{\dfrac{21}{12}}{\dfrac{4}{9}} = \frac{21}{12} \times \frac{9}{4} = \frac{189}{48} = \frac{63}{16}

Since α,β∈(0,π2)\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right) and \tan(\alpha+\beta) > 0, we have α+β∈(0,π2)\alpha + \beta \in \left(0, \dfrac{\pi}{2}\right).

Therefore: α+β=tan⁡−16316\alpha + \beta = \tan^{-1}\dfrac{63}{16}, i.e., tan⁡−16316=sin⁡−1513+cos⁡−135\tan^{-1}\dfrac{63}{16} = \sin^{-1}\dfrac{5}{13} + \cos^{-1}\dfrac{3}{5}. Hence Proved.\qquad \textbf{Hence Proved.}
8Prove that tan⁡−1x=12cos⁡−11−x1+x\tan^{-1}\sqrt{x} = \dfrac{1}{2}\cos^{-1}\dfrac{1-x}{1+x}, x∈[0,1]x \in [0,1].Show solution
To Prove: tan⁡−1x=12cos⁡−11−x1+x\tan^{-1}\sqrt{x} = \dfrac{1}{2}\cos^{-1}\dfrac{1-x}{1+x}

Proof:

Let x=tan⁡θ\sqrt{x} = \tan\theta, so x=tan⁡2θx = \tan^2\theta and θ=tan⁡−1x\theta = \tan^{-1}\sqrt{x}.

Since x∈[0,1]x \in [0,1], we have θ∈[0,π4]\theta \in \left[0, \dfrac{\pi}{4}\right], so 2θ∈[0,π2]⊂[0,π]2\theta \in \left[0, \dfrac{\pi}{2}\right] \subset [0, \pi].

Now:
1−x1+x=1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\frac{1-x}{1+x} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta

Since 2θ∈[0,π]2\theta \in [0, \pi]:
cos⁡−1(1−x1+x)=cos⁡−1(cos⁡2θ)=2θ\cos^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}(\cos 2\theta) = 2\theta

Therefore:
12cos⁡−11−x1+x=θ=tan⁡−1xHence Proved.\frac{1}{2}\cos^{-1}\frac{1-x}{1+x} = \theta = \tan^{-1}\sqrt{x} \qquad \textbf{Hence Proved.}
9Prove that cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)=x2\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) = \dfrac{x}{2}, x∈(0,π4)x \in \left(0, \dfrac{\pi}{4}\right).Show solution
To Prove: cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)=x2\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) = \dfrac{x}{2}

Proof:

Using half-angle identities:
1+sin⁡x=(cos⁡x2+sin⁡x2)2=cos⁡x2+sin⁡x2\sqrt{1+\sin x} = \sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2} = \cos\frac{x}{2}+\sin\frac{x}{2}
1−sin⁡x=(cos⁡x2−sin⁡x2)2=cos⁡x2−sin⁡x2\sqrt{1-\sin x} = \sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2} = \cos\frac{x}{2}-\sin\frac{x}{2}

(Both are positive since x∈(0,π4)x \in \left(0, \dfrac{\pi}{4}\right) implies x2∈(0,π8)\dfrac{x}{2} \in \left(0, \dfrac{\pi}{8}\right), so \cos\dfrac{x}{2} > \sin\dfrac{x}{2} > 0.)

Substituting:
1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x=(cos⁡x2+sin⁡x2)+(cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)−(cos⁡x2−sin⁡x2)=2cos⁡x22sin⁡x2=cot⁡x2\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} = \frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)+\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)-\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)} = \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}

Therefore:
cot⁡−1(cot⁡x2)=x2\cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}

(Since x2∈(0,π8)⊂(0,π)\dfrac{x}{2} \in \left(0, \dfrac{\pi}{8}\right) \subset (0, \pi), which is the range of cot⁡−1\cot^{-1}.)

Hence Proved.\textbf{Hence Proved.}
10Prove that tan⁡−1(1+x−1−x1+x+1−x)=π4−12cos⁡−1x\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}x, −12≤x≤1-\dfrac{1}{\sqrt{2}} \leq x \leq 1. [Hint: Put x=cos⁡2θx = \cos 2\theta]Show solution
To Prove: tan⁡−1(1+x−1−x1+x+1−x)=π4−12cos⁡−1x\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}x

Proof:

Let x=cos⁡2θx = \cos 2\theta, so 2θ=cos⁡−1x2\theta = \cos^{-1}x, i.e., θ=12cos⁡−1x\theta = \dfrac{1}{2}\cos^{-1}x.

Since x∈[−12,1]x \in \left[-\dfrac{1}{\sqrt{2}}, 1\right], we have 2θ∈[0,3π4]2\theta \in \left[0, \dfrac{3\pi}{4}\right], so θ∈[0,3π8]\theta \in \left[0, \dfrac{3\pi}{8}\right].

Now:
1+x=1+cos⁡2θ=2cos⁡2θ=2cos⁡θ\sqrt{1+x} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta
1−x=1−cos⁡2θ=2sin⁡2θ=2sin⁡θ\sqrt{1-x} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta

(Both positive since θ∈[0,3π8]\theta \in \left[0, \dfrac{3\pi}{8}\right].)

Substituting:
1+x−1−x1+x+1−x=2cos⁡θ−2sin⁡θ2cos⁡θ+2sin⁡θ=cos⁡θ−sin⁡θcos⁡θ+sin⁡θ=1−tan⁡θ1+tan⁡θ=tan⁡(π4−θ)\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} = \frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)

Therefore:
tan⁡−1(tan⁡(π4−θ))=π4−θ=π4−12cos⁡−1x\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x

Hence Proved.\textbf{Hence Proved.}
11Solve: 2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2\tan^{-1}(\cos x) = \tan^{-1}(2\operatorname{cosec}x).Show solution
Given: 2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2\tan^{-1}(\cos x) = \tan^{-1}(2\operatorname{cosec}x)

Working:

Using the identity 2tan⁡−1a=tan⁡−12a1−a22\tan^{-1}a = \tan^{-1}\dfrac{2a}{1-a^2} (for |a| < 1):
tan⁡−12cos⁡x1−cos⁡2x=tan⁡−1(2cosec⁡x)\tan^{-1}\frac{2\cos x}{1 - \cos^2 x} = \tan^{-1}(2\operatorname{cosec}x)

2cos⁡xsin⁡2x=2sin⁡x\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}

2cos⁡xsin⁡2x=2sin⁡x\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}

Multiplying both sides by sin⁡2x\sin^2 x:
2cos⁡x=2sin⁡x2\cos x = 2\sin x

tan⁡x=1\tan x = 1

x=π4x = \frac{\pi}{4}

Answer: x=π4x = \dfrac{\pi}{4}.
12Solve: tan⁡−11−x1+x=12tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x} = \dfrac{1}{2}\tan^{-1}x, (x > 0).Show solution
Given: tan⁡−11−x1+x=12tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x} = \dfrac{1}{2}\tan^{-1}x, x > 0

Working:

Using the identity tan⁡−1a−tan⁡−1b=tan⁡−1a−b1+ab\tan^{-1}a - \tan^{-1}b = \tan^{-1}\dfrac{a-b}{1+ab}:
tan⁡−11−x1+x=tan⁡−11−tan⁡−1x=π4−tan⁡−1x\tan^{-1}\frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x = \frac{\pi}{4} - \tan^{-1}x

So the equation becomes:
π4−tan⁡−1x=12tan⁡−1x\frac{\pi}{4} - \tan^{-1}x = \frac{1}{2}\tan^{-1}x

π4=32tan⁡−1x\frac{\pi}{4} = \frac{3}{2}\tan^{-1}x

tan⁡−1x=π6\tan^{-1}x = \frac{\pi}{6}

x=tan⁡π6=13x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}

Answer: x=13x = \dfrac{1}{\sqrt{3}}.
13sin⁡(tan⁡−1x)\sin(\tan^{-1}x), |x| < 1 is equal to
(A) x1−x2\dfrac{x}{\sqrt{1-x^2}}
(B) 11−x2\dfrac{1}{\sqrt{1-x^2}}
(C) 11+x2\dfrac{1}{\sqrt{1+x^2}}
(D) x1+x2\dfrac{x}{\sqrt{1+x^2}}Show solution
Correct Option: (D) x1+x2\dfrac{x}{\sqrt{1+x^2}}

Working:

Let tan⁡−1x=θ\tan^{-1}x = \theta, so tan⁡θ=x\tan\theta = x.

From the right triangle with opposite side xx and adjacent side 11, the hypotenuse is 1+x2\sqrt{1+x^2}.

Therefore:
sin⁡θ=x1+x2\sin\theta = \frac{x}{\sqrt{1+x^2}}

Hence sin⁡(tan⁡−1x)=x1+x2\sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1+x^2}}.

Answer: Option (D) x1+x2\dfrac{x}{\sqrt{1+x^2}}.
14sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x) - 2\sin^{-1}x = \dfrac{\pi}{2}, then xx is equal to
(A) 0,120, \dfrac{1}{2}
(B) 1,121, \dfrac{1}{2}
(C) 00
(D) 12\dfrac{1}{2}Show solution
Correct Option: (C) 00

Working:

Given: sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x) - 2\sin^{-1}x = \dfrac{\pi}{2}

sin⁡−1(1−x)=π2+2sin⁡−1x\sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1}x

Applying sin⁡\sin to both sides:
1−x=sin⁡(π2+2sin⁡−1x)=cos⁡(2sin⁡−1x)1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}x\right) = \cos(2\sin^{-1}x)

Let sin⁡−1x=θ\sin^{-1}x = \theta, so sin⁡θ=x\sin\theta = x:
1−x=cos⁡2θ=1−2sin⁡2θ=1−2x21 - x = \cos 2\theta = 1 - 2\sin^2\theta = 1 - 2x^2

1−x=1−2x21 - x = 1 - 2x^2

2x2−x=02x^2 - x = 0

x(2x−1)=0x(2x - 1) = 0

x=0orx=12x = 0 \quad \text{or} \quad x = \frac{1}{2}

Verification:
- For x=0x = 0: sin⁡−1(1)−0=π2\sin^{-1}(1) - 0 = \dfrac{\pi}{2} ✓
- For x=12x = \dfrac{1}{2}: sin⁡−1(12)−2sin⁡−1(12)=π6−π3=−π6≠π2\sin^{-1}\left(\dfrac{1}{2}\right) - 2\sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6} - \dfrac{\pi}{3} = -\dfrac{\pi}{6} \neq \dfrac{\pi}{2} ✗

So x=12x = \dfrac{1}{2} does not satisfy the original equation.

Answer: Option (C) x=0x = 0.

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