Let x=tan^2θ⇒θ =tan^ 1√(x) Now, 1/2cos^ 1 ( 1 x/1+x ) =1/2cos^ 1 ( 1 tan^2θ/1+tan^2θ ) =1/2cos^ 1cos2θ=2θ/2=θ=tan^ 1√(x). ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Composition of Inverse Trigonometric Functions and Trigonometric Functions

1/2cos-1 1-x/...

Question

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

t a n^{- 1} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

c o t^{- 1} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $t a n^{- 1} \sqrt{x}$
Let $x = t a n^{2} θ \Rightarrow θ = t a n^{- 1} \sqrt{x}$
Now, $\frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x})$
$= \frac{1}{2} c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})$
$= \frac{1}{2} c o s^{- 1} c o s 2 θ = \frac{2 θ}{2} = θ = t a n^{- 1} \sqrt{x}$ .

Suggest Corrections

Similar questions

$t a n^{- 1} \sqrt{x} = \frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x}), x ϵ [0, 1] .$

\frac{1}{2} {cos}^{- 1} (\frac{1 - x}{1 + x}) = {tan}^{- 1} \sqrt{x}

{tan}^{- 1} \sqrt{x} = \frac{1}{2} {cos}^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1]

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

x = {tan}^{- 1} 1 - {cos}^{- 1} [- \frac{1}{2}] + {sin}^{- 1} \frac{1}{2}; y = cos [\frac{1}{2} {cos}^{- 1} [\frac{1}{8}]]

Join BYJU'S Learning Program