#### Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 15

$f(x) \text { is decreasing function on } (\frac{\pi }{4},\frac{\pi }{2})$

Given:

$f(x) =tan^{-1}(sin\: x+cos\: x)$

To prove:

$\text { We have to show that } f(x) \text { is decreasing function on } (\frac{\pi }{4},\frac{\pi }{2})$

Hint:

Condition for decreasing function f’(x)<0.

Solution:

Given

$f(x) =tan^{-1}(sin\: x+cos\: x)$

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\tan ^{-1}(\sin x+\cos x)\right) \\ &\Rightarrow f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^{2}} \times(\cos x-\sin x),\left[\therefore \frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}\right] \\ &\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \\ &\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{2(1+\sin x \cos x)},\left[\therefore \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}

Now, as given

\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}

\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ \end{aligned}

as here cosine values are smaller than sine values for same angle.

\begin{aligned} &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}

Hence condition for f(x) to be decreasing.

$\text { Thus }f(x) \text { is decreasing on interval } x\: \in \: (\frac{\pi }{4},\frac{\pi }{2})$