Question

(cos^2 A+1)tan^2 A ÷sin^2 A =1/cos^2 A solve this

Answer 1

Given:-

$\dfrac{(cos^2A+1) tan^2A }{Sin^2A} = \dfrac{1}{Cos^2A}$

Considering L. H. S

$\dfrac{(cos^2A+1) tan^2A }{Sin^2A}$

We know that $tanA = \dfrac{sinA}{cosA}$

put the value of tanA.

=$\dfrac{(Cos^2A+1)\dfrac{Sin^2A}{Cos^2A}}{Sin^2A}$

=$\dfrac{(\cancel{cos^2A}\times\dfrac{Sin^2A}{\cancel{Cos^2A}})+(\dfrac{Sin^2A}{Cos^2A})}{Sin^2A}$

=$\dfrac{Sin^2A+\dfrac{Sin^2A}{Cos^2A}}{Sin^2A}$

=$\dfrac{\dfrac{Sin^2ACos^2A+Sin^2A}{Cos^2A}}{Sin^2A}$

=$\dfrac{\cancel{Sin^2ACos^2A}}{\cancel{Sin^2ACos^2A}}+\dfrac{\cancel{Sin^2A}}{\cancel{Sin^2A}Cos^2A}$

$1 + \dfrac{1}{Cos^2A}$

L. H. S = R. H. S

hence,proved