cot−1[(cosα)12]−tan−1[(cosα)12]=x, then sinx=

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c o t^{- 1} [(c o s α)^{\frac{1}{2}}] - t a n^{- 1} [(c o s α)^{\frac{1}{2}}] = x

, then

s i n x =

Options:

A . tan2(α2)

B . cot2(α2)

C . tan α

D . cot(α2)

Answer: Option A
:
A

t a n^{- 1} [\frac{1}{\sqrt{c o s α}}] - t a n^{- 1} [\sqrt{c o s α}] = x

\Rightarrow t a n^{- 1} ⎡ ⎢ ⎣ \frac{\frac{1}{\sqrt{c o s α}} - \sqrt{c o s α}}{1 + \frac{\sqrt{c o s α}}{\sqrt{c o s α}}} ⎤ ⎥ ⎦ = x \Rightarrow t a n x = \frac{1 - c o s α}{2 \sqrt{c o s α}}

∴ s i n x = \frac{1 - c o s α}{1 + c o s α} = \frac{2 s i n^{2} \frac{α}{2}}{2 c o s^{2} \frac{α}{2}} = t a n^{2} (\frac{α}{2})

.

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