### Inverse Trigonometric Functions Question 8

#### Question 8 - 01 February - Shift 2

Let $S={x \in R: 0<x<1.$ and $.2 \tan ^{-1}(\frac{1-x}{1+x})=\cos ^{-1}(\frac{1-x^{2}}{1+x^{2}})}$. If $n(S)$ denotes the number of elements in $S$ then :

(1) $n(S)=2$ and only one element in $S$ is less then $\frac{1}{2}$.

(2) $n(S)=1$ and the element in $S$ is more than $\frac{1}{2}$.

(3) $n(S)=1$ and the element in $S$ is less than $\frac{1}{2}$.

(4) $n(S)=0$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Formula of subtraction of tangent function, Formula of double angle of cosine function in tan form

$0<x<1$

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

$\tan ^{-1} x=\theta \in(0, \frac{\pi}{4}) \therefore x=\tan \theta$

$2 \tan ^{-1}(\tan (\frac{\pi}{4}-\theta))=\cos ^{-1}(\cos 2 \theta)$

$2(\frac{\pi}{4}-\theta)=2 \theta \quad \therefore 4 \theta=\frac{\pi}{2} \quad \therefore \theta=\frac{\pi}{8}$

$x=\tan \frac{\pi}{8} \quad \therefore x=\sqrt{2}-1 \simeq 0.414$