Question

Proove the identity: (SecA-tanA)(tanA+secA)/(cosecA-cotA)= cotA+cosecA

Answer 1

\(\large\tt \color{black}{\dfrac{(secA-tanA)(tanA+secA)}{cosecA-cotA}}\)


\(\large\tt \color{black}{=\dfrac{sec^2A-tan^2A}{cosecA-cotA}}\)


\(\large\tt \color{black}{=\dfrac{1}{cosecA-cotA}~~~(as~~sec^2A-tan^2A=1)}\)


\(\large\tt \color{black}{=\dfrac{1}{cosecA-cotA}\times\dfrac{cosecA+cotA}{cosecA+cotA}}\)


\(\large\tt \color{black}{=\dfrac{cosecA+cotA}{cosec^2A-cot^2A}}\)


\(\large\tt \color{black}{=cosecA+cotA~~~(as~~cosec^2A-cot^A=1)}\)