Question

$\tan ^{2} ( \alpha ) = \cos ^{2} ( \beta) - \sin ^{2} \beta then \: prove \: that\: \cos ^{2} ( \alpha ) - \sin ^{2} ( \alpha ) = \tan ^{2} ( \beta )$
pls answer these question guys​

Answer 1

Step-by-step explanation:

We have,

$\begin{aligned} \\ & \qquad { \tan }^{2} ( \alpha ) = { \cos }^{2} ( \beta ) - { \sin }^{2} ( \beta ) \\ \\ & \implies \frac{ { \sin }^{2} ( \alpha )}{ { \cos }^{2} ( \alpha )} = { \cos}^{2}( \beta ) - { \sin}^{2}( \beta ) \\ \\ & \implies \frac{ { \sin }^{2}( \alpha ) }{ { \cos }^{2}( \alpha ) { \cos }^{2}( \beta ) } = \frac{ \cancel{{ \cos}^{2}( \beta)} }{ \cancel{{ \cos }^{2} ( \beta) }} - \frac{ { \sin }^{2}( \beta ) }{ { \cos}^{2}( \beta ) } \\ &\qquad \qquad \longrightarrow \left[ \sf dividing \: throughout \: by \: { \cos }^{2}( \beta ) \right] \\ \\ & \implies \frac{ { \sin }^{2}( \alpha ) }{ { \cos }^{2}( \alpha ) { \cos }^{2}( \beta ) } = 1 - { \tan }^{2} ( \beta ) \\ \\ & \implies { \tan }^{2}( \beta ) = 1 - \frac{ { \sin }^{2}( \alpha ) }{ { \cos }^{2}( \alpha ) { \cos }^{2}( \beta ) } \\ \\ & \qquad \qquad \qquad = \frac{{ { \cos }^{2}( \alpha ) { \cos }^{2}( \beta ) } - { \sin }^{2}( \alpha ) }{ { \cos }^{2}( \alpha ) { \cos }^{2}( \beta ) } \end{aligned}$