Short Question 15 – 18

Question 15:

$Prove the identity \frac{2}{\cos 2 A + 1} = s e c^{2} A$

Solution:

$\begin{array}{l} LHS \\ = \frac{2}{\cos 2 A + 1} \\ = \frac{2}{(2 \cos^{2} A - 1) + 1} \leftarrow \cos 2 A = 2 \cos^{2} A - 1 \\ = \frac{2}{2 \cos^{2} A} = \frac{1}{\cos^{2} A} \\ = s e c^{2} A \\ = RHS \\ ∴ Proven \end{array}$

Question 16:

$Prove the identity \frac{2 \tan A}{2 - s e c^{2} A} = \tan 2 A$

Solution:

$\begin{array}{l} LHS \\ = \frac{2 \tan A}{2 - s e c^{2} A} \\ = \frac{2 \tan A}{2 - (\tan^{2} A + 1)} \leftarrow \tan^{2} A + 1 = s e c^{2} A \\ = \frac{2 \tan A}{1 - \tan^{2} A} \\ = \tan 2 A \\ = RHS \\ ∴ Proven \end{array}$

Question 17:

$Prove the identity \tan x + \cot x = 2 \cos e c 2 x$

Solution:

$\begin{array}{l} LHS \\ = \tan x + \cot x \\ = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \\ = \frac{\sin^{2} x + \cos^{2} x}{\cos x \sin x} \\ = \frac{1}{\cos x \sin x} \leftarrow \sin^{2} x + \cos^{2} x = 1 \\ = \frac{1}{\frac{1}{2} \sin 2 x} \leftarrow \begin{array}{l} \sin 2 x = 2 \sin x \cos x \\ \frac{1}{2} \sin 2 x = \sin x \cos x \end{array} \\ = \frac{2}{\sin 2 x} \\ = 2 (\frac{1}{\sin 2 x}) \\ = 2 \cos e c 2 x \\ = RHS \\ ∴ Proven \end{array}$

Question 18:

$Prove the identity \frac{\cos x - \sin 2 x}{\cos 2 x + \sin x - 1} = \frac{1}{\tan x}$

Solution:

$\begin{array}{l} LHS \\ = \frac{\cos x - \sin 2 x}{\cos 2 x + \sin x - 1} \\ = \frac{\cos x - 2 \sin x \cos x}{(1 - 2 \sin^{2} x) + \sin x - 1} \leftarrow \cos 2 x = 1 - 2 \sin^{2} x \\ = \frac{\cos x (1 - 2 \sin x)}{\sin x - 2 \sin^{2} x} \\ = \frac{\cos x (1 - 2 \sin x)}{\sin x (1 - 2 \sin x)} \\ = \frac{\cos x}{\sin x} \\ = \cot x \\ = \frac{1}{\tan x} \\ = RHS \\ ∴ Proven \end{array}$