Question

If y=tan-1(sinx+cosx/cosx-sinx) then find dy/dx.

Answer 1

i hope it will help you.

Answer 2

Answer:

$\dfrac{dy}{dx}\ =\ 1$

Step-by-step explanation:

As given in question,

•      $y\ =\ \tan^{-1}(\dfrac{sinx+cosx}{cosx-sinx})$

$=>\ tany\ =\ \dfrac{cosx+sinx}{cosx-sinx}$

Divide both numerator and denominator of RHS by cosx, we will have

$=>\ tany\ =\ \dfrac{1+tanx}{1-tanx}$

$=>\ tany\ =\ \dfrac{tan\dfrac{\pi}{4}+tanx}{1-tan\dfrac{\pi}{4}.tanx}$

$=>\ tany\ =\ tan(\dfrac{\pi}{4}+x)$

$=>\ y\ =\ \dfrac{\pi}{4}+x$

Now, by differentiating on both sides of the equation with respect to x, we will have

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

                       = 1

Hence, for the given value of y the value of $\dfrac{dy}{dx}$ will be 1.