Question

Evaluate ${\tan }^{-1}\left[\frac{\cos \mathrm{x}}{1+\sin \mathrm{x}}\right]$

• A. $\left(\frac{\mathrm{\pi }}{4}\right)-\left(\frac{\mathrm{x}}{2}\right)$
• B. $\left(\frac{\mathrm{\pi }}{4}\right)+\left(\frac{\mathrm{x}}{2}\right)$
• C. $\frac{\mathrm{x}}{2}$
• D. $\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{x}$

Answer 1

The correct option is A

$\left(\frac{\mathrm{\pi }}{4}\right)-\left(\frac{\mathrm{x}}{2}\right)$

Explanation for correct option

Evaluating the given expression:

Given,

${\tan }^{-1}\left[\frac{\cos \mathrm{x}}{1+\sin \mathrm{x}}\right]$

Let's equate the expression:

${\tan }^{-1}\left[\frac{\mathrm{cosx}}{1+\mathrm{sinx}}\right]={\tan }^{-1}\left[\frac{\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}{1+\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}\right]\left[\because \sin \left(\frac{\pi }{2}-\theta \right)=\mathrm{cos\theta }\right]$

We know that,

$\mathbf{sin}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{2}\mathbf{sin\theta cos\theta }$ and $\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\theta }$

Therefore,

$\mathbf{sin\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{sin}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{cos}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ and $\mathbf{1}\mathbf{+}\mathbf{cos\theta }\mathbf{=}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}....\left(2\right)$

Putting $\mathrm{\theta }=\frac{\mathrm{\pi }}{2}-\mathrm{x}$ in $\left(1\right)$:

$\Rightarrow \sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=2\sin \frac{1}{2}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)\cos \frac{1}{2}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)$

$=2\sin \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{x}}{2}\right)\cos \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{x}}{2}\right)$

Putting $\mathrm{\theta }=\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)$ in $\left(2\right)$:

$1+\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=2{\cos }^2\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)$

$=2{\cos }^2\left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{x}}{2}\right)$

Hence, the expression becomes,

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

$={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{x}}{2}\right)\right]$

$=\left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{x}}{2}\right)$

Therefore, the answer is $\mathrm{(}\frac{\mathrm{\pi }}{\mathrm{4}}\mathrm{-}\frac{\mathrm{x}}{\mathrm{2}}\mathrm{)}$.

Hence, the correct answer is option (A).