Question

( tanA + 1/ cosA )² + ( tanA - 1/cosA )² =
2( 1 + sin²A/ 1 - sin²A )​

Answer 1

Step-by-step explanation:

lhs α+cosα1)2+(cosαsinα−cosα1)2=cos2α2(sin2α+1)

$\begin{gathered} {( \tan \alpha + \frac{1}{ \cos \alpha } )}^{2} +( \tan \alpha - \frac{1}{ \cos \alpha }) { }^{2} \\ = ( \frac{ \sin\alpha }{ \cos \alpha } + \frac{1}{ \cos\alpha } ) {}^{2} + ( \frac{ \sin \alpha }{ \cos\alpha } - \frac{1}{ \cos\alpha }) {}^{2} \\ = \frac{2( \sin {}^{2} \alpha + 1) }{ \cos {}^{2} \alpha } \end{gathered}(tanα+cosα1)2+(tanα−cosα1)2=(cosαsinα+cosα1)2+(cosαsinα−cosα1)2=cos2α2(sin2α+1)$

RHS

2 (1+ sin²α)/(1- sin²α)

2 ( sin²α+1)/cos²α

LHS = RHS