Question

sec^2theta by cosec theta​

Answer 1

Answer:

sec2θ+cosec2θ=cos2θ1+sin2θ1=sin2θ

Answer 2

Correct question:-

prove that Sec²A+Cosec²A = Sec²A.Cosec²A

Solution:-

L.H.S = Sec²A+Cosec²A

$\implies$ $\dfrac{1}{Cos²A}$ + $\dfrac{1}{Sin²A}$

$\implies$ $\dfrac{Sin²A+Cos²A}{Cos²A.Sin²A}$

$\implies$ $\dfrac{1}{Cos²A.Sin²A}$ = Sec²A.Cosec²A = R.H.S

hence, verified.