Question

The value of $\sin \left[{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$1$

Explanation for the correct option

Step 1. Find ${\tan }^{-1}$ in terms of ${\sin }^{-1}$

The given trigonometric expression: $\sin \left[{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$.

Let us assume that, ${\tan }^{-1}\left(\frac{1-x^2}{2x}\right)=y$.

$$\begin{gathered} \Rightarrow \tan \left(y\right)=\frac{1-x^2}{2x} \\ \Rightarrow {\tan }^2\left(y\right)={\left[\frac{1-x^2}{2x}\right]}^2 \\ \Rightarrow {\sec }^2\left(\mathrm{y}\right)-1=\frac{1+{\mathrm{x}}^4-2{\mathrm{x}}^2}{4{\mathrm{x}}^2}\left[\because {\sec }^2\left(\mathrm{y}\right)-{\tan }^2\left(\mathrm{y}\right)=1\right] \\ \Rightarrow {\sec }^2\left(\mathrm{y}\right)=\frac{1+{\mathrm{x}}^4-2{\mathrm{x}}^2}{4{\mathrm{x}}^2}+1 \\ \Rightarrow {\sec }^2\left(\mathrm{y}\right)=\frac{1+{\mathrm{x}}^4-2{\mathrm{x}}^2+4{\mathrm{x}}^2}{4{\mathrm{x}}^2} \\ \Rightarrow {\sec }^2\left(\mathrm{y}\right)=\frac{1+{\mathrm{x}}^4+2{\mathrm{x}}^2}{4{\mathrm{x}}^2} \\ \Rightarrow {\sec }^2\left(\mathrm{y}\right)=\frac{{\left(1+{\mathrm{x}}^2\right)}^2}{4{\mathrm{x}}^2} \\ \Rightarrow {\cos }^2\left(\mathrm{y}\right)=\frac{4{\mathrm{x}}^2}{{\left(1+{\mathrm{x}}^2\right)}^2}\left[\sec \left(y\right)=\frac{1}{\cos \left(y\right)}\right] \\ \Rightarrow 1-{\sin }^2\left(\mathrm{y}\right)=\frac{4{\mathrm{x}}^2}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ \Rightarrow {\sin }^2\left(\mathrm{y}\right)=1-\frac{4{\mathrm{x}}^2}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ \Rightarrow {\sin }^2\left(\mathrm{y}\right)=\frac{{\left(1+{\mathrm{x}}^2\right)}^2-4{\mathrm{x}}^2}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ \Rightarrow {\sin }^2\left(\mathrm{y}\right)=\frac{{\left(1-{\mathrm{x}}^2\right)}^2}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ \Rightarrow \sin \left(\mathrm{y}\right)=\frac{\left(1-{\mathrm{x}}^2\right)}{\left(1+{\mathrm{x}}^2\right)} \\ \Rightarrow \mathrm{y}={\sin }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right) \\ \Rightarrow {\tan }^{-1}\left(\frac{1-{\mathrm{x}}^2}{2\mathrm{x}}\right)={\sin }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right) \end{gathered}$$

Step 2. Find the value of the expression

Thus, $\sin \left[{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]=\sin \left[{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$.

$$\begin{gathered} =\sin \left[\frac{\mathrm{\pi }}{2}\right]\left[{\sin }^{-1}\left(\theta \right)+{\cos }^{-1}\left(\theta \right)=\frac{\pi }{2}\right] \\ =1 \end{gathered}$$

Therefore, the value of $\sin \left[{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ is $1$.

Hence, the correct option is (A).