Question

Differentiate with respect to x
cot^-1 ( cosecx+ cotx)

Answer 1

Answer:

Step-by-step explanation:

$cosec^{2}x-cot^{2}x=1\\(cosecx-cotx)(cosecx+cotx)=1\\(cosecx+cotx)=\frac{1}{(cosecx-cotx)}\\Also\\Tan^{-1}x=Cot^{-1}(\frac{1}{x})\\Now\\Cot^{-1}(cosecx+cotx)\\=Cot^{-1}(\frac{1}{cosecx-cotx})\\=Tan^{-1}(cosecx+cotx)\\Now \\f(x)=Tan^{-1}(cosecx+cotx)\\WKT \:\:\\\frac{d}{dx}(Tan^{-1}x)=\frac{1}{1+x^2}\\\frac{d}{dx}(cosecx)=-cosecx \:cotx\\\frac{d}{dx}(cotx)=-cosec^{2}x\\Now\\f^{I}(x)=\frac{1}{1+(cosecx+cotx)^{2}}\times(-cosecx \:cotx-cosec^{2}x)\\f^{I}(x)=\frac{-cosecx(cosecx+cotx)}{1+(cosecx+cotx)^{2}}$
contd in attachment

Hope it helps