c o t − 1 [ ( c o s α ) 1 2 ] − t a n − 1 [ ( c o s α ) 1 2 ] = x , then s i n x = Options: A . &nbsp t a n 2 ( α 2 ) B . &nbsp c o t 2 ( α 2 ) C . &nbsp t a n α D . &nbsp c o t ( α 2 )

Answer: Option A : A t a n − 1 [ 1 √ c o s α ] − t a n − 1 [ √ c o s α ] = x ⇒ t a n − 1 ⎡ ⎢ ⎣ 1 √ c o s α − √ c o s α 1 + √ c o s α √ c o s α ⎤ ⎥ ⎦ = x ⇒ t a n x = 1 − c o s α 2 √ c o s α ∴ s i n x = 1 − c o s α 1 + c o s α = 2 s i n 2 α 2 2 c o s 2 α 2 = t a n 2 ( α 2 ) .

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

Discussion Forum : Inverse Trigonometric Functions

Question -

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

Options:

A .

t a n^{2} (\frac{α}{2})

B .

c o t^{2} (\frac{α}{2})

C .

t a n α

D .

c o t (\frac{α}{2})

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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Discussion Forum : Inverse Trigonometric Functions

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