6.7.5 Trigonometric Functions Short Questions (Question 11 - 14)

Question 11:

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

Solution:

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

Question 12:

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

Solution:

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

Question 13:

$Prove the identity \tan^{2} θ - \sin^{2} θ = \tan^{2} θ \sin^{2} θ$

Solution:

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

Question 14:

${Prove the identity cosec}^{2} θ (\sec^{2} θ - \tan^{2} θ) - 1 = \cot^{2} θ$

Solution:

$\begin{array}{l} LHS \\ = {cosec}^{2} θ (\sec^{2} θ - \tan^{2} θ) - 1 \\ = {cosec}^{2} θ (1) - 1 \leftarrow \begin{array}{l} \tan^{2} θ + 1 = \sec^{2} θ \\ \sec^{2} θ - \tan^{2} θ = 1 \end{array} \\ = {cosec}^{2} θ - 1 \\ = \cot^{2} θ \leftarrow \begin{array}{l} 1 + \cot^{2} θ = \cos e c^{2} θ \\ \cos e c^{2} θ - 1 = \cot^{2} θ \end{array} \\ = RHS \\ ∴ Proven \end{array}$

6.7.5 Trigonometric Functions Short Questions (Question 11 – 14)

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