First slide

Multiple and sub- multiple Angles

Question

Which of the following identities, wherever defined hold(s) good?

Moderate

A

$\cot α - \tan α = 2 \cot 2 α$
B

$\tan (45^{\circ} + α) - \tan (45^{\circ} - α) = 2 cosec 2 α$
C

$\tan (45^{\circ} + α) + \tan (45^{\circ} - α) = 2 \sec 2 α$
D

$\tan α + \cot α = 2 \cdot \tan 2 α$

Solution

$(a) \frac{\cos^{2} α - \sin^{2} α}{\sin αcos α} = 2 \cot 2 α$
$\begin{array}{l} (b) \frac{1 + t}{1 - t} - \frac{1 - t}{1 + t}, where t = \tan α \\ = \frac{(1 + t)^{2} - (1 - t)^{2}}{1 - t^{2}} \\ = \frac{4 t}{1 - t^{2}} = \frac{2 \times 2 \tan α}{1 - \tan^{2} α} = 2 \tan 2 α \end{array}$
Hence, (b) is incorrect.
$\begin{array}{l} (c) \frac{1 + t}{1 - t} + \frac{1 - t}{1 + t} = \frac{(1 + t)^{2} + (1 - t)^{2}}{1 - t^{2}} \\ = \frac{2 (1 + t^{2})}{1 - t^{2}} (where t - tanα) \\ = \frac{2}{\cos 2 α} = 2 \sec 2 α \end{array}$
Hence, (c) is correct.
$(d) \tan α + \cot α = \frac{1}{\cos αsin α} = \frac{2}{\sin 2 α} = 2 cosec 2 α,$
Hence, (d) is incorrect.

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