Introduction to Trigonometry

Given: △ABC right-angled at B, AB = 24 cm, BC = 7 cm.

Step 1: Find the hypotenuse AC.

By Pythagoras theorem:
$$AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625$$
$$AC = 25 \text{ cm}$$

(i) For angle A:
- Side opposite to A = BC = 7 cm
- Side adjacent to A = AB = 24 cm
- Hypotenuse = AC = 25 cm

$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$$

$$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$$

(ii) For angle C:
- Side opposite to C = AB = 24 cm
- Side adjacent to C = BC = 7 cm
- Hypotenuse = AC = 25 cm

$$\sin C = \frac{AB}{AC} = \frac{24}{25}$$

$$\cos C = \frac{BC}{AC} = \frac{7}{25}$$

Note: From the figure, △PQR is right-angled at Q with PQ = 12 cm and PR = 13 cm.

Step 1: Find QR using Pythagoras theorem.
$$PR^2 = PQ^2 + QR^2$$
$$13^2 = 12^2 + QR^2$$
$$169 = 144 + QR^2$$
$$QR^2 = 25 \implies QR = 5 \text{ cm}$$

Step 2: Calculate tan P and cot R.

$$\tan P = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{PQ} = \frac{5}{12}$$

$$\cot R = \frac{\text{adjacent}}{\text{opposite}} = \frac{QR}{PQ} = \frac{5}{12}$$

(For angle R: opposite = PQ = 12, adjacent = QR = 5, so $\tan R = \frac{12}{5}$ and $\cot R = \frac{5}{12}$.)

Step 3:
$$\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = \boxed{0}$$

Given: $\sin A = \dfrac{3}{4}$

Step 1: Use the identity $\sin^2 A + \cos^2 A = 1$.
$$\cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}$$
$$\cos A = \frac{\sqrt{7}}{4}$$

(Taking positive value since A is acute.)

Step 2: Calculate tan A.
$$\tan A = \frac{\sin A}{\cos A} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$$

Answer: $\cos A = \dfrac{\sqrt{7}}{4}$, $\tan A = \dfrac{3}{\sqrt{7}} = \dfrac{3\sqrt{7}}{7}$

Given: $15\cot A = 8 \implies \cot A = \dfrac{8}{15}$

Step 1: Construct a right triangle.

Let the side adjacent to A = 8k and side opposite to A = 15k.

Hypotenuse $= \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$

Step 2: Find sin A and sec A.
$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15k}{17k} = \frac{15}{17}$$

$$\sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{17k}{8k} = \frac{17}{8}$$

Answer: $\sin A = \dfrac{15}{17}$, $\sec A = \dfrac{17}{8}$

Given: $\sec\theta = \dfrac{13}{12}$, so $\cos\theta = \dfrac{12}{13}$.

Step 1: Construct a right triangle.

Hypotenuse = 13k, adjacent side = 12k.

Opposite side $= \sqrt{(13k)^2 - (12k)^2} = \sqrt{169k^2 - 144k^2} = \sqrt{25k^2} = 5k$

Step 2: Write all ratios.

$$\sin\theta = \frac{5k}{13k} = \frac{5}{13}$$

$$\cos\theta = \frac{12}{13}$$

$$\tan\theta = \frac{5k}{12k} = \frac{5}{12}$$

$$\cot\theta = \frac{12}{5}$$

$$\sec\theta = \frac{13}{12}$$

$$\csc\theta = \frac{13}{5}$$

Given: ∠A and ∠B are acute angles and $\cos A = \cos B$.

To prove: $\angle A = \angle B$

Proof:

Consider a right triangle. Let $\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}$.

Suppose the triangles have the same hypotenuse (or we can use the ratio argument):

Let $\cos A = \dfrac{AC_1}{H}$ and $\cos B = \dfrac{AC_2}{H}$ for the same hypotenuse $H$.

Since $\cos A = \cos B$:
$$\frac{AC_1}{H} = \frac{AC_2}{H} \implies AC_1 = AC_2$$

This means the adjacent sides are equal, so the triangles are congruent (by RHS), which gives $\angle A = \angle B$.

Alternatively (using the fact that cosine is a one-to-one function for acute angles):

For acute angles, $\cos\theta$ is a strictly decreasing function. Therefore, if $\cos A = \cos B$ and both A, B are acute, then:
$$\angle A = \angle B \quad \blacksquare$$

Given: $\cot\theta = \dfrac{7}{8}$

Step 1: Find sin θ and cos θ.

Opposite = 8k, Adjacent = 7k, Hypotenuse $= \sqrt{64k^2+49k^2} = \sqrt{113}\,k$

$$\sin\theta = \frac{8}{\sqrt{113}}, \quad \cos\theta = \frac{7}{\sqrt{113}}$$

(i) Using the identity $(1-\sin^2\theta) = \cos^2\theta$ and $(1-\cos^2\theta) = \sin^2\theta$:

$$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta$$

$$= \left(\frac{7}{8}\right)^2 = \frac{49}{64}$$

(ii)
$$\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$$

Given: $3\cot A = 4 \implies \cot A = \dfrac{4}{3} \implies \tan A = \dfrac{3}{4}$

Step 1: Find sin A and cos A.

Opposite = 3k, Adjacent = 4k, Hypotenuse = 5k.

$$\sin A = \frac{3}{5}, \quad \cos A = \frac{4}{5}$$

Step 2: Calculate LHS.
$$\text{LHS} = \frac{1-\tan^2 A}{1+\tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$$

Step 3: Calculate RHS.
$$\text{RHS} = \cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$

Conclusion: LHS = RHS = $\dfrac{7}{25}$. ✓ The equation holds true.

Given: △ABC right-angled at B, $\tan A = \dfrac{1}{\sqrt{3}}$.

Step 1: Find angle A.
$$\tan A = \frac{1}{\sqrt{3}} = \tan 30^\circ \implies \angle A = 30^\circ$$

Since ∠B = 90°, we have $\angle C = 180° - 90° - 30° = 60°$.

Step 2: Write the values.
$$\sin A = \sin 30^\circ = \frac{1}{2}, \quad \cos A = \cos 30^\circ = \frac{\sqrt{3}}{2}$$
$$\sin C = \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos C = \cos 60^\circ = \frac{1}{2}$$

(i)
$$\sin A\cos C + \cos A\sin C = \frac{1}{2}\cdot\frac{1}{2} + \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1$$

(ii)
$$\cos A\cos C - \sin A\sin C = \frac{\sqrt{3}}{2}\cdot\frac{1}{2} - \frac{1}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$$

Given: △PQR right-angled at Q, PR + QR = 25 cm, PQ = 5 cm.

Step 1: Let QR = x, then PR = 25 – x.

By Pythagoras theorem:
$$PR^2 = PQ^2 + QR^2$$
$$(25-x)^2 = 5^2 + x^2$$
$$625 - 50x + x^2 = 25 + x^2$$
$$625 - 50x = 25$$
$$50x = 600 \implies x = 12$$

So $QR = 12$ cm and $PR = 25 - 12 = 13$ cm.

Step 2: Calculate the ratios.

$$\sin P = \frac{QR}{PR} = \frac{12}{13}$$

$$\cos P = \frac{PQ}{PR} = \frac{5}{13}$$

$$\tan P = \frac{QR}{PQ} = \frac{12}{5}$$

(i) False.
$\tan A$ is the ratio of the opposite side to the adjacent side. For example, \tan 60^\circ = \sqrt{3} > 1. So tan A can be greater than, equal to, or less than 1. The statement is False.

(ii) True.
$\sec A = \dfrac{\text{hypotenuse}}{\text{adjacent side}} \geq 1$ for all acute angles. Since \dfrac{12}{5} = 2.4 > 1, it is possible for some angle A. The statement is True.

(iii) False.
$\cos A$ is the abbreviation for cosine of angle A, not cosecant. Cosecant is abbreviated as $\csc A$ (or $\text{cosec } A$). The statement is False.

(iv) False.
$\cot A$ is a single trigonometric ratio (cotangent of angle A). It is not the product of some quantity 'cot' and 'A'. The statement is False.

(v) False.
The value of $\sin\theta$ always lies between –1 and 1 (i.e., $-1 \leq \sin\theta \leq 1$). Since \dfrac{4}{3} > 1, $\sin\theta = \dfrac{4}{3}$ is not possible for any angle $\theta$. The statement is False.

Standard values used:
$$\sin 30^\circ = \frac{1}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2},\ \sin 60^\circ = \frac{\sqrt{3}}{2},\ \cos 60^\circ = \frac{1}{2}$$
$$\tan 45^\circ = 1,\ \sec 30^\circ = \frac{2}{\sqrt{3}},\ \csc 30^\circ = 2,\ \cos 45^\circ = \frac{1}{\sqrt{2}}$$
$$\cot 45^\circ = 1,\ \csc 60^\circ = \frac{2}{\sqrt{3}}$$

(i)
$$\sin 60^\circ\cos 30^\circ + \sin 30^\circ\cos 60^\circ = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4} + \frac{1}{4} = \boxed{1}$$

(ii)
$$2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 = 2 + \frac{3}{4} - \frac{3}{4} = \boxed{2}$$

(iii)
$$\frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}} + 2} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2(1+\sqrt{3})}$$
$$= \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} \times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)} = \frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}(3-1)} = \frac{3-\sqrt{3}}{4\sqrt{2}} = \frac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}$$
$$= \frac{3-\sqrt{3}}{4\sqrt{2}} = \frac{(3-\sqrt{3})\sqrt{2}}{8} = \boxed{\dfrac{3\sqrt{2}-\sqrt{6}}{8}}$$

(iv)
$$\frac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} = \frac{\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} + \dfrac{1}{2} + 1}$$

Numerator: $\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}} = \dfrac{3}{2} - \dfrac{2}{\sqrt{3}} = \dfrac{3\sqrt{3} - 4}{2\sqrt{3}}$

Denominator: $\dfrac{2}{\sqrt{3}} + \dfrac{3}{2} = \dfrac{4 + 3\sqrt{3}}{2\sqrt{3}}$

$$= \frac{3\sqrt{3}-4}{4+3\sqrt{3}} \times \frac{4-3\sqrt{3}}{4-3\sqrt{3}} = \frac{(3\sqrt{3}-4)(4-3\sqrt{3})}{16-27} = \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11} = \frac{24\sqrt{3}-43}{-11} = \boxed{\dfrac{43-24\sqrt{3}}{11}}$$

(v)
$$\frac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$$

Denominator $= \sin^2 30^\circ + \cos^2 30^\circ = 1$ (Pythagorean identity)

Numerator $= 5\left(\dfrac{1}{2}\right)^2 + 4\left(\dfrac{2}{\sqrt{3}}\right)^2 - (1)^2 = 5\cdot\dfrac{1}{4} + 4\cdot\dfrac{4}{3} - 1 = \dfrac{5}{4} + \dfrac{16}{3} - 1$

$$= \frac{15}{12} + \frac{64}{12} - \frac{12}{12} = \frac{67}{12}$$

$$\text{Answer} = \frac{67/12}{1} = \boxed{\dfrac{67}{12}}$$

(i) $\dfrac{2\tan 30^\circ}{1+\tan^2 30^\circ}$

$$= \frac{2 \cdot \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}} = \frac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2} = \sin 60^\circ$$

Correct option: (A) $\sin 60^\circ$

(ii) $\dfrac{1-\tan^2 45^\circ}{1+\tan^2 45^\circ}$

$$= \frac{1 - 1^2}{1 + 1^2} = \frac{0}{2} = 0$$

Correct option: (D) 0

(iii) $\sin 2A = 2\sin A$

$$2\sin A\cos A = 2\sin A \implies 2\sin A(\cos A - 1) = 0$$

This holds when $\sin A = 0$, i.e., $A = 0^\circ$.

Correct option: (A) $0^\circ$

(iv) $\dfrac{2\tan 30^\circ}{1-\tan^2 30^\circ}$

$$= \frac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} = \frac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ$$

Correct option: (C) $\tan 60^\circ$

Given: $\tan(A+B) = \sqrt{3}$ and $\tan(A-B) = \dfrac{1}{\sqrt{3}}$

Step 1: From $\tan(A+B) = \sqrt{3} = \tan 60^\circ$:
$$A + B = 60^\circ \quad \cdots (1)$$

Step 2: From $\tan(A-B) = \dfrac{1}{\sqrt{3}} = \tan 30^\circ$:
$$A - B = 30^\circ \quad \cdots (2)$$

Step 3: Adding (1) and (2):
$$2A = 90^\circ \implies A = 45^\circ$$

Step 4: Subtracting (2) from (1):
$$2B = 30^\circ \implies B = 15^\circ$$

Answer: $A = 45^\circ$, $B = 15^\circ$

(i) False.
Counterexample: Let A = B = 30°.
$\sin(30°+30°) = \sin 60° = \dfrac{\sqrt{3}}{2}$
$\sin 30° + \sin 30° = \dfrac{1}{2} + \dfrac{1}{2} = 1 \neq \dfrac{\sqrt{3}}{2}$. False.

(ii) True.
From the table: $\sin 0^\circ = 0,\ \sin 30^\circ = 0.5,\ \sin 45^\circ \approx 0.707,\ \sin 60^\circ \approx 0.866,\ \sin 90^\circ = 1$. The value increases as θ increases from 0° to 90°. True.

(iii) False.
From the table: $\cos 0^\circ = 1,\ \cos 30^\circ \approx 0.866,\ \cos 45^\circ \approx 0.707,\ \cos 60^\circ = 0.5,\ \cos 90^\circ = 0$. The value decreases as θ increases. False.

(iv) False.
$\sin\theta = \cos\theta$ only when $\theta = 45^\circ$. For example, $\sin 30^\circ = 0.5 \neq \cos 30^\circ = \dfrac{\sqrt{3}}{2}$. False.

(v) True.
$\cot A = \dfrac{\cos A}{\sin A}$. At $A = 0^\circ$, $\sin 0^\circ = 0$, so $\cot 0^\circ = \dfrac{1}{0}$, which is undefined. True.

Using fundamental identities:

(i) sin A in terms of cot A:

We know $\csc^2 A = 1 + \cot^2 A$, so $\sin^2 A = \dfrac{1}{\csc^2 A} = \dfrac{1}{1+\cot^2 A}$

$$\boxed{\sin A = \frac{1}{\sqrt{1+\cot^2 A}}}$$

(positive since A is acute)

(ii) sec A in terms of cot A:

$\sec^2 A = 1 + \tan^2 A = 1 + \dfrac{1}{\cot^2 A} = \dfrac{\cot^2 A + 1}{\cot^2 A}$

$$\boxed{\sec A = \frac{\sqrt{1+\cot^2 A}}{\cot A}}$$

(iii) tan A in terms of cot A:

$$\boxed{\tan A = \frac{1}{\cot A}}$$

Given: sec A (so cos A = 1/sec A is known).

sin A: Using $\sin^2 A + \cos^2 A = 1$:
$$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$$

cos A:
$$\cos A = \frac{1}{\sec A}$$

tan A: Using $\sec^2 A - \tan^2 A = 1$:
$$\tan A = \sqrt{\sec^2 A - 1}$$

cot A:
$$\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}$$

csc A:
$$\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$$

(i) $9\sec^2 A - 9\tan^2 A = 9(\sec^2 A - \tan^2 A) = 9 \times 1 = 9$

Correct option: (B) 9

(ii) $(1+\tan\theta+\sec\theta)(1+\cot\theta-\csc\theta)$

$$= \left(1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta}\right)\left(1 + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}\right)$$

$$= \frac{\cos\theta + \sin\theta + 1}{\cos\theta} \cdot \frac{\sin\theta + \cos\theta - 1}{\sin\theta}$$

$$= \frac{(\sin\theta+\cos\theta)^2 - 1}{\sin\theta\cos\theta} = \frac{\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta - 1}{\sin\theta\cos\theta}$$

$$= \frac{1 + 2\sin\theta\cos\theta - 1}{\sin\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2$$

Correct option: (C) 2

(iii) $(\sec A + \tan A)(1 - \sin A)$

$$= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1-\sin A) = \frac{1+\sin A}{\cos A}\cdot(1-\sin A)$$

$$= \frac{1-\sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$$

Correct option: (D) cos A

(iv) $\dfrac{1+\tan^2 A}{1+\cot^2 A} = \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Correct option: (D) $\tan^2 A$

LHS: Write in terms of $\sin\theta$ and $\cos\theta$. Let $s = \sin\theta$, $c = \cos\theta$.

$$\frac{s/c}{1 - c/s} + \frac{c/s}{1 - s/c} = \frac{s/c}{(s-c)/s} + \frac{c/s}{(c-s)/c}$$

$$= \frac{s^2}{c(s-c)} + \frac{c^2}{s(c-s)} = \frac{s^2}{c(s-c)} - \frac{c^2}{s(s-c)}$$

$$= \frac{1}{s-c}\left(\frac{s^2}{c} - \frac{c^2}{s}\right) = \frac{1}{s-c}\cdot\frac{s^3 - c^3}{sc}$$

$$= \frac{(s-c)(s^2+sc+c^2)}{sc(s-c)} = \frac{s^2+sc+c^2}{sc}$$

$$= \frac{\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta}{\sin\theta\cos\theta}$$

$$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} + 1 = \sec\theta\csc\theta + 1 = \text{RHS}$$

$$\blacksquare$$

LHS:
$$\frac{1+\sec A}{\sec A} = \frac{1 + \dfrac{1}{\cos A}}{\dfrac{1}{\cos A}} = \frac{\dfrac{\cos A + 1}{\cos A}}{\dfrac{1}{\cos A}} = \cos A + 1 = 1 + \cos A$$

RHS:
$$\frac{\sin^2 A}{1-\cos A} = \frac{1-\cos^2 A}{1-\cos A} = \frac{(1-\cos A)(1+\cos A)}{1-\cos A} = 1 + \cos A$$

LHS = RHS = $1 + \cos A$ $\blacksquare$

Divide numerator and denominator by $\sin A$:

$$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}$$

Substitute $1 = \csc^2 A - \cot^2 A$ in the denominator:

Denominator $= \cot A - \csc A + 1 = \cot A - \csc A + \csc^2 A - \cot^2 A$
$= (\csc A - \cot A)(\csc A + \cot A) - (\csc A - \cot A)$
$= (\csc A - \cot A)(\csc A + \cot A - 1)$

Numerator $= \cot A + \csc A - 1$

$$\text{LHS} = \frac{\csc A + \cot A - 1}{(\csc A - \cot A)(\csc A + \cot A - 1)} = \frac{1}{\csc A - \cot A}$$

$$= \frac{1}{\csc A - \cot A} \times \frac{\csc A + \cot A}{\csc A + \cot A} = \frac{\csc A + \cot A}{\csc^2 A - \cot^2 A} = \frac{\csc A + \cot A}{1} = \csc A + \cot A = \text{RHS}$$

$$\blacksquare$$

LHS:
$$\frac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)}$$

Note: $1 - 2\sin^2\theta = 1 - 2(1-\cos^2\theta) = 2\cos^2\theta - 1$

$$= \frac{\sin\theta(2\cos^2\theta - 1)}{\cos\theta(2\cos^2\theta - 1)} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}$$

$$\blacksquare$$

LHS:
$$(\csc A - \sin A)(\sec A - \cos A) = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right)$$

$$= \frac{1-\sin^2 A}{\sin A} \cdot \frac{1-\cos^2 A}{\cos A} = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \sin A\cos A$$

RHS:
$$\frac{1}{\tan A + \cot A} = \frac{1}{\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}} = \frac{1}{\dfrac{\sin^2 A + \cos^2 A}{\sin A\cos A}} = \frac{\sin A\cos A}{1} = \sin A\cos A$$

LHS = RHS = $\sin A\cos A$ $\blacksquare$

Part 1: Show $\dfrac{1+\tan^2 A}{1+\cot^2 A} = \tan^2 A$

$$\frac{1+\tan^2 A}{1+\cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \checkmark$$

Part 2: Show $\left(\dfrac{1-\tan A}{1-\cot A}\right)^2 = \tan^2 A$

$$\frac{1-\tan A}{1-\cot A} = \frac{1 - \dfrac{\sin A}{\cos A}}{1 - \dfrac{\cos A}{\sin A}} = \frac{\dfrac{\cos A - \sin A}{\cos A}}{\dfrac{\sin A - \cos A}{\sin A}}$$

$$= \frac{(\cos A - \sin A)}{\cos A} \cdot \frac{\sin A}{(\sin A - \cos A)} = \frac{-(\sin A - \cos A)\sin A}{\cos A(\sin A - \cos A)} = \frac{-\sin A}{\cos A} = -\tan A$$

$$\left(\frac{1-\tan A}{1-\cot A}\right)^2 = (-\tan A)^2 = \tan^2 A \checkmark$$

Hence all three expressions are equal to $\tan^2 A$. $\blacksquare$