Question

sin square theta + sec square theta is equal to sin square theta into cos square theta + 1 upon cos squared theta​

Answer 1

$\huge\mathcal{Hey\:mate}$

To prove =

${ \sin( \theta) }^{2} + { \sec( \theta) }^{2} = \frac{ { \sin( \theta) }^{2} . { \cos( \theta) }^{2} + 1}{ { \cos( \theta) }^{2} }$

We can solve it by both LHS and RHS

Answer refers to the attachment ❤

Answer 2

Given

${ \sin( \alpha ) }^{2} + { \sec( \alpha ) }^{2} = \frac{ { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} + 1 }{ { \cos( \alpha ) }^{2} }$

Taking LHS

${ \sin( \alpha ) }^{2} + { \sec( \alpha ) }^{2} \\ { \sin( \alpha ) }^{2} + \frac{1}{ { \cos( \alpha ) }^{2} } \\ \frac{ { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} + 1 }{ { \cos( \alpha ) }^{2} }$

Hence LHS =RHS

proved