Q:

prove d/dx(tanx/(1+secx))=csc²x-cotxcscx

Accepted Solution

A:
First, let's find the derivative:[tex]f(x)=\dfrac{tan\ x}{1+sec\ x}\\\\\\f'(x)=\dfrac{a'b-ab'}{b^2}\\\\\rightarrow a=tan\ x\qquad \qquad a'=\sec^2\ x\\\rightarrow b=1+sec\ x\qquad \ b'=sec\x \cdot tan\ x\\\\\\f'(x)=\dfrac{sec^2\ x(1+sec\ x)-tan\ x(sec\ x\cdot tan\ x)}{(1+sec\ x)^2}\\\\\\.\quad =\dfrac{1}{1+cos\ x}[/tex]Next, manipulate the derivative to match the right hand side:[tex]\dfrac{1}{1+cos\ x}\bigg(\dfrac{1-cos\ x}{1-cos\ x}\bigg)\\\\\\\rightarrow \quad \dfrac{1-cos\ x}{1-cos^2\ x}\\\\\\\rightarrow \quad \dfrac{1-cos\ x}{sin^2\ x}\\\\\\\rightarrow \quad \dfrac{1}{sin^2\ x}- \dfrac{cos\ x}{sin^2\ x}\\\\\\\rightarrow \quad \dfrac{1}{sin^2\ x}- \dfrac{cos\ x}{sin\ x}\cdot \dfrac{1}{sin\ x}\\\\\\\rightarrow \quad csc^2\ x-cot\ x\cdot csc\ x[/tex]LHS = RHS [tex]\checkmark[/tex]