Question

Diffrentiate: $\frac{\sin (ax+b)}{\cos (cx+d)}$

Answer 1

Given:Let $y=\frac{\sin (ax+b)}{\cos (cx+d)}$

Let $u=\sin (ax+b)$ and $v=\cos (cx+d)$

Therefore,

$\frac{du}{dx}=\cos (ax+b)\times \frac{d(ax+b)}{dx}$

$=\cos (ax+b)\times a$ ...(i)

And

$\frac{dv}{dx}=-\sin (cx+d)\times \frac{d(cx+d)}{dx}$

$=-\sin (cx+d)\times c$...(ii)

Now,

$\frac{dy}{dx}=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

From (i) and (ii)

$=\frac{a\cos (ax+b)\cdot \cos (cx+d)+c\sin (cx+d)\cdot \sin (ax+b)}{\left(\cos \right(cx+d){)}^2}$

$=\frac{a\cos (ax+b)\cdot \cos (cx+d)}{{\cos }^2(cx+d)}+\frac{c\sin (cx+d)\cdot \sin (ax+b)}{{\cos }^2(cx+d)}$

$=a\cos (ax+b)\frac{1}{cos(cx+d)}+c\sin (ax+b)\frac{\sin (cx+d)}{cos(cx+d)}\times \frac{1}{cos(cx+d)}$

$=a\cos (ax+b)\cdot \sec (cx+d)+c\sin (ax+b)\cdot \tan (cx+d)\cdot \sec (cx+d)$