Inverse Trigonometric Functions — NCERT Solutions

First evaluate the inside angle:

$$\sin^{-1}\frac12=\frac{\pi}{6}$$

So,

$$2\sin^{-1}\frac12=2\cdot\frac{\pi}{6}=\frac{\pi}{3}$$

Then

$$2\cos\left(\frac{\pi}{3}\right)=2\cdot\frac12=1$$

Therefore,

$$\tan^{-1}[2\cos(2\sin^{-1}\tfrac12)] = \tan^{-1}(1)=\frac{\pi}{4}$$

So the value is $\frac{\pi}{4}$.

Using the standard identities from the chapter:

$$\sin^{-1}\left(\frac{2x}{1+x^2}\right)=2\tan^{-1}x$$
for the given valid range, and

$$\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)=2\tan^{-1}y$$
for $y>0$.

Hence

$$\frac12\left[\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right] =\tan^{-1}x+\tan^{-1}y$$

Therefore,

$$\tan\left(\frac12[\cdots]\right)=\tan(\tan^{-1}x+\tan^{-1}y) =\frac{x+y}{1-xy}$$

So the simplified form is $\frac{x+y}{1-xy}$.

$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$$
The principal value range of $\sin^{-1}$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Now,

$$\sin\frac{2\pi}{3}=\sin\left(\pi-\frac{2\pi}{3}\right)=\sin\frac{\pi}{3}$$
and $\frac{\pi}{3}$ lies in the principal range, so

$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right)=\frac{\pi}{3}$$

The principal value range of $\tan^{-1}$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Since

$$\tan\frac{3\pi}{4}=-1$$

we get

$$\tan^{-1}\left(\tan\frac{3\pi}{4}\right)=\tan^{-1}(-1)=-\frac{\pi}{4}$$

because $-\frac{\pi}{4}$ lies in the principal range.

Let

$$A=\sin^{-1}\frac35,\qquad B=\cot^{-1}\frac32$$

From $\sin A=\frac35$, in a right triangle take opposite $=3$, hypotenuse $=5$, so adjacent $=4$. Thus

$$\tan A=\frac34$$

For $B$, $\cot B=\frac32$, so

$$\tan B=\frac23$$

Now

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} =\frac{\frac34+\frac23}{1-\frac34\cdot\frac23}$$

$$=\frac{\frac{9+8}{12}}{1-\frac12} =\frac{\frac{17}{12}}{\frac12} =\frac{17}{6}$$

So the value is $\frac{17}{6}$.

The principal value range of $\cos^{-1}x$ is $[0,\pi]$. Since
$$\cos\frac{7\pi}{6}=\cos\left(\pi+\frac{\pi}{6}\right)=-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2},$$
we need the angle in $[0,\pi]$ whose cosine is $-\frac{\sqrt3}{2}$. That angle is
$$\cos^{-1}\left(\cos\frac{7\pi}{6}\right)=\frac{5\pi}{6}.$$

$$\sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}$$
because the principal value range of $\sin^{-1}$ is $\left[-\frac\pi2,\frac\pi2\right]$.
So,
$$\sin\left(\frac\pi3-\sin^{-1}\left(-\frac12\right)\right) =\sin\left(\frac\pi3+\frac\pi6\right) =\sin\frac\pi2 =1.$$
The computed value is $1$, which matches option (D).

The principal values are:
$$\tan^{-1}\sqrt3=\frac\pi3$$
since $\tan\frac\pi3=\sqrt3$, and
$$\cot^{-1}(-\sqrt3)=\frac{5\pi}{6}$$
because the principal value range of $\cot^{-1}$ is $(0,\pi)$ and $\cot\frac{5\pi}{6}=-\sqrt3$.
Hence,
$$\tan^{-1}\sqrt3-\cot^{-1}(-\sqrt3) =\frac\pi3-\frac{5\pi}{6} =-\frac\pi2.$$
So the correct option is (B).

$$\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$$
Now

$$\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$$
so

$$\cos\frac{13\pi}{6}=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$$
The principal value range of $\cos^{-1}$ is $[0,\pi]$, hence

$$\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\cos^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{6}$$

The principal value range of $\tan^{-1}$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Since

$$\tan\frac{7\pi}{6}=\tan\left(\pi+\frac{\pi}{6}\right)=\tan\frac{\pi}{6}=\frac{1}{\sqrt3}$$

But the angle $\frac{7\pi}{6}$ is in the third quadrant, so the tangent value is positive. Therefore,

$$\tan^{-1}\left(\tan\frac{7\pi}{6}\right)=\tan^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}$$

So the value is $\frac{\pi}{6}$.

Let $\sin^{-1}\frac35=A$ and $\cot^{-1}\frac32=B$.

Then
$$\sin A=\frac35 \implies \cos A=\frac45 \implies \tan A=\frac34.$$
Also,
$$\cot B=\frac32 \implies \tan B=\frac23.$$
Now use
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} = \frac{\frac34+\frac23}{1-\frac34\cdot\frac23} = \frac{\frac{17}{12}}{\frac12} =\frac{17}{6}.$$
So the value of the expression is $\frac{17}{6}$.

Let
$$A=\sin^{-1}\frac{8}{17},\qquad B=\sin^{-1}\frac35.$$
Then
$$\sin A=\frac{8}{17},\quad \cos A=\frac{15}{17}$$
because $8^2+15^2=17^2$, and
$$\sin B=\frac35, \quad \cos B=\frac45.$$
Now,
$$\sin(A+B)=\sin A\cos B+\cos A\sin B =\frac{8}{17}\cdot\frac45+\frac{15}{17}\cdot\frac35 =\frac{32}{85}+\frac{45}{85} =\frac{77}{85}.$$
Also,
$$\cos(A+B)=\cos A\cos B-\sin A\sin B =\frac{15}{17}\cdot\frac45-\frac{8}{17}\cdot\frac35 =\frac{60-24}{85} =\frac{36}{85}.$$
Therefore,
$$\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}=\frac{77/85}{36/85}=\frac{77}{36}.$$
Hence,
$$\sin^{-1}\frac{8}{17}+\sin^{-1}\frac35=\tan^{-1}\frac{77}{36}.$$

Let
$$A=\cos^{-1}\frac45,\qquad B=\cos^{-1}\frac{12}{13}.$$
Then
$$\cos A=\frac45,\quad \sin A=\frac35,$$
$$\cos B=\frac{12}{13},\quad \sin B=\frac{5}{13}.$$
Now,
$$\cos(A+B)=\cos A\cos B-\sin A\sin B =\frac45\cdot\frac{12}{13}-\frac35\cdot\frac{5}{13} =\frac{48-15}{65} =\frac{33}{65}.$$
Since the principal value of $\cos^{-1}$ lies in $[0,\pi]$, the sum equals
$$\cos^{-1}\frac{33}{65}.$$
So the identity is proved.

Let
$$A=\cos^{-1}\frac{12}{13},\qquad B=\sin^{-1}\frac35.$$
Then
$$\cos A=\frac{12}{13},\quad \sin A=\frac{5}{13},$$
$$\sin B=\frac35, \quad \cos B=\frac45.$$
Now,
$$\sin(A+B)=\sin A\cos B+\cos A\sin B =\frac{5}{13}\cdot\frac45+\frac{12}{13}\cdot\frac35 =\frac{20}{65}+\frac{36}{65} =\frac{56}{65}.$$
Hence,
$$\cos^{-1}\frac{12}{13}+\sin^{-1}\frac35=\sin^{-1}\frac{56}{65}.$$

Let
$$A=\sin^{-1}\frac{5}{13},\qquad B=\cos^{-1}\frac35.$$
Then
$$\sin A=\frac{5}{13},\quad \cos A=\frac{12}{13},\quad \tan A=\frac{5}{12},$$
and
$$\cos B=\frac35,\quad \sin B=\frac45,\quad \tan B=\frac43.$$
So
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} =\frac{\frac{5}{12}+\frac43}{1-\frac{5}{12}\cdot\frac43} =\frac{\frac{21}{12}}{1-\frac59} =\frac{\frac74}{\frac49} =\frac{63}{16}.$$
Therefore,
$$\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac35.$$

Let
$$\theta=\tan^{-1}\sqrt{x}.$$
Then
$$\tan\theta=\sqrt{x}.$$
Using the identity
$$\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta},$$
we get
$$\cos 2\theta=\frac{1-x}{1+x}.$$
Hence,
$$2\theta=\cos^{-1}\left(\frac{1-x}{1+x}\right)$$
for $x\in[0,1]$. Therefore,
$$\tan^{-1}\sqrt{x}=\frac12\cos^{-1}\left(\frac{1-x}{1+x}\right).$$

Let
$$A=\sqrt{1+\sin x},\qquad B=\sqrt{1-\sin x}.$$
Then the expression becomes
$$\cot^{-1}\left(\frac{A+B}{A-B}\right).$$
Multiply numerator and denominator inside by $A+B$:
$$\frac{A+B}{A-B}=\frac{(A+B)^2}{A^2-B^2}.$$
Now,
$$A^2-B^2=(1+\sin x)-(1-\sin x)=2\sin x,$$
and
$$(A+B)^2=A^2+B^2+2AB=2+2\sqrt{1-\sin^2x}=2+2\cos x.$$
So
$$\frac{A+B}{A-B}=\frac{2(1+\cos x)}{2\sin x}=\frac{1+\cos x}{\sin x}.$$
Using half-angle identities,
$$\frac{1+\cos x}{\sin x}=\cot\frac{x}{2}.$$
Therefore,
$$\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}$$
for the given range of $x$, where $\frac{x}{2}$ lies in the principal branch of $\cot^{-1}$.

Let
$$x=\cos 2\theta.$$
Then for $-\frac{1}{\sqrt2}\le x\le 1$, we may take $0\le \theta\le \frac{3\pi}{4}$, and
$$\sqrt{1+x}=\sqrt{1+\cos 2\theta}=\sqrt{2\cos^2\theta}=\sqrt2\cos\theta,$$
$$\sqrt{1-x}=\sqrt{1-\cos 2\theta}=\sqrt{2\sin^2\theta}=\sqrt2\sin\theta$$
(with the appropriate signs in the given interval).
Thus
$$\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} =\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta} =\tan\left(\frac\pi4-\theta\right).$$
Hence
$$\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) =\frac\pi4-\theta.$$
Since $x=\cos 2\theta$, we have $\theta=\frac12\cos^{-1}x$, so
$$\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) =\frac\pi4-\frac12\cos^{-1}x.$$