6.7.6 Trigonometric Functions Short Questions (Question 15 - 18)

Question 15:

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

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

Solution:

\begin{matrix} 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{matrix}

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

Question 16:

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

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

Solution:

\begin{matrix} 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{matrix}

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

Question 17:

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

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

Solution:

\begin{matrix} 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{matrix} sin 2 x = 2 sin x cos x \frac{1}{2} sin 2 x = sin x cos x \end{matrix} = \frac{2}{sin 2 x} = 2 (\frac{1}{sin 2 x}) = 2 cos e c 2 x = RHS ∴ Proven \end{matrix}

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

Question 18:

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

Solution:

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

6.7.6 Trigonometric Functions Short Questions (Question 15 – 18)

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