1 2 c o s − 1 ( 1 − x 1 + x ) = Options: A . &nbsp c o t − 1 √ x B . &nbsp t a n − 1 √ x C . &nbsp t a n − 1 x D . &nbsp c o t − 1 x

Answer: Option B : B Let x = t a n 2 θ ⇒ θ = t a n − 1 √ x Now, 1 2 c o s − 1 ( 1 − x 1 + x ) = 1 2 c o s − 1 ( 1 − t a n 2 θ 1 + t a n 2 θ ) = 1 2 c o s − 1 c o s 2 θ = 2 θ 2 = θ = t a n − 1 √ x .

12cos−1(1−x1+x)=

Discussion Forum : Inverse Trigonometric Functions

Question -

$\frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x}) =$

Options:

A .

c o t^{- 1} \sqrt{x}

B .

t a n^{- 1} \sqrt{x}

C .

t a n^{- 1} x

D .

c o t^{- 1} x

Answer: Option B
:
B
Let

x = t a n^{2} θ \Rightarrow θ = t a n^{- 1} \sqrt{x}

Now,

\frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x})

= \frac{1}{2} c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})

= \frac{1}{2} c o s^{- 1} c o s 2 θ = \frac{2 θ}{2} = θ = t a n^{- 1} \sqrt{x}

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