Question

solve one of them( the one with correct answer be given 20 thnx)​

Answer 1

Answer:

This is your answer. Hope it helps.

Answer 2

★ Rationale : -

→ To prove

${ \sec \theta }^{2} - \: {cos \: \theta}^{2} = { \sin\theta}^{2} \: ( { \sec\theta }^{2} + 1)$

★ LHS :

${ \sec \theta }^{2} - { \cos\theta }^{2}$

$( \sec \theta - \cos \theta) \: ( \sec \theta + \cos \theta)$

$( \dfrac{1}{ \cos\theta } - \cos\theta) \: ( \dfrac{1}{ \cos\theta } + \cos\theta)$

$( \dfrac{1 - { \cos }\theta^{2} }{ \cos\theta} )( \dfrac{1 + { \cos }\theta^{2} }{ \cos\theta} )$

$= ( \dfrac{ { \sin } ^ {2} \theta}{ \cos \theta } )( \dfrac{1}{ \cos \: \theta } + \dfrac{ { \cos \: }^{2}\theta }{ \cos \theta } )$

$= { \sin }^{2} \theta ( \dfrac{1}{ { \cos }^{2} \theta } + 1)$

$= { \sin}^{2} \theta \: ( { \sec } ^{2} \theta \: + 1)$

= RHS

Hence proved.