Question

Question ✌️

the function f(x)
$= \tan {}^{ - 1} ( \sin(x) + \cos(x) )$
is an increasing function in interval ?
​

Answer 1

Given :

Function :

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

To find :

Increasing interval of given function .

Solution :

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

Differentiate it with respect to x

$\sf f'(x) = \dfrac{1}{1 + ( \sin x + \cos x) {}^{2} } \times ( \cos x - \sin x)$

$= \sf \dfrac{ \sqrt{2} ( \frac{1}{ \sqrt{2} } \cos x - \frac{1}{ \sqrt{2} } \sin x) }{1 +( \sin x + \cos x) {}^{2} }$

$= \sf \dfrac{ \sqrt{2} ( \cos \frac{\pi}{4} \cos x - \sin x \sin \frac{\pi}{4} )}{1 + ( \sin x + \cos x) {}^{2} }$

$= \sf \dfrac{ \sqrt{2} \times \cos(x + \frac{\pi}{4} )}{1 + ( \sin x + \cos x) {}^{2} }$

If f'(x) >0 then f(x) is increasing function .

Hence ,f(x) is increasing , if

$\frac{ - \pi}{2 } < x + \frac{\pi}{4} < \frac{\pi}{2}$

$\implies \frac{ - 3\pi}{4} < x < \frac{\pi}{4}$

Therefore , increasing interval of given function is $[\dfrac{-\pi}{2},\dfrac{-\pi}{4}]$

Answer 2

Answer:

is this anisha ?????....

Step-by-step explanation:

the phenomenal