Answer to Question #92144 in Calculus for Noblemundackal

Question

Differentiate tan^-1((sins-Cosx)/(Sini Cox’s))

Answer 1

$$(tan^{-1}(\frac{sin(x)-cos(x)}{sin(x)cos(x)}))'=$$

$$(tan^{-1}(\frac{1}{cos(x)}-\frac{1}{sin(x)}))'=$$

(using chain rule)

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

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

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

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

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

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

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

$$\frac{sin^2(x)cos^2(x)}{(1-sin(x)cos(x))^2}\frac{sin^3(x)+cos^3(x)}{sin^2(x)cos^2(x)}=$$

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

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

$$\frac{sin(x)+cos(x)}{1-sin(x)cos(x)}$$

Answer:

$$\frac{sin(x)+cos(x)}{1-sin(x)cos(x)}$$