Question

If $f\left(x\right)=\frac{{\sin }^{2}x}{1+\cot x}+\frac{{\cos }^{2}x}{1+\tan x}$, then $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$ is equal to

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $0$
• D. $-\sqrt{3}$

Answer 1

The correct option is C

$0$

Explanation for the correct option.

Step 1: Simplifying the given function:

$\begin{array}{rcl}f\left(x\right)& =& \frac{{\sin }^{2}x}{1+\cot x}+\frac{{\cos }^{2}x}{1+\tan x}\\ & =& \frac{{\sin }^{2}x}{1+{\displaystyle \frac{1}{\tan x}}}+\frac{{\cos }^{2}x}{1+\tan x}\\ & =& \frac{{\displaystyle \frac{\sin x}{\cos x}}{\sin }^{2}x}{1+\tan x}+\frac{{\cos }^{2}x}{1+\tan x}\\ & =& \frac{{\sin }^{3}x+{\cos }^{3}x}{\cos x\left(1+\tan x\right)}\\ & =& \frac{{\sin }^{3}x+{\cos }^{3}x}{\cos x+\sin x}\\ & =& \frac{\left(\sin x+\cos x\right)\left({\sin }^{2}x+{\cos }^{2}x-\sin x·\cos x\right)}{\cos x+\sin x}\mathbf{}\mathbf{\left[}\mathbf{\because }{\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{a}\mathbf{b}\mathbf{)}\mathbf{\right]}\\ & =& 1-\sin x·\cos x\\ & =& 1-\frac{1}{2}\sin 2x\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathit{x}\mathbf{c}\mathbf{o}\mathbf{s}\mathit{x}\mathbf{]}\end{array}$

Step 2: Find the value of $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$:

Now, differentiate $f\left(x\right)$ w.r.t. $x$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& -\cos 2x\\ & \Rightarrow & f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=-\cos 2\left(\frac{\mathrm{\pi }}{4}\right)\\ & =& -\cos \frac{\mathrm{\pi }}{2}\\ & =& 0\end{array}$

Hence, the correct option is C.