If y = tan-1( sinx+cosx/cox−sinx) then dy/dx is equal to

The correct option is 1.

To find the derivative of the given function, we first simplify the expression inside the inverse tangent function:

$y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$

We divide both the numerator and the denominator of the fraction inside the parentheses by $\cos x$:

$\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}=\frac{\tan x+1}{1-\tan x}=\frac{1+\tan x}{1-\tan x}$

Using the trigonometric identity $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ and substituting $\tan \left(\frac{\pi }{4}\right)=1$, we can rewrite the expression as:

$\frac{1+\tan x}{1-\tan x}=\tan (\frac{\pi }{4}+x)$

Now, substitute this back into the original function:

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

Since the inverse function cancels the trigonometric function, we get:

$y=\frac{\pi }{4}+x$

Next, we differentiate $y$ with respect to $x$ using the sum rule of differentiation:

$\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi }{4}+x)$

Since $\frac{\pi }{4}$ is a constant, its derivative is $0$, and the derivative of $x$ with respect to $x$ is $1$:

$\frac{dy}{dx}=0+1=1$