Question

Prove that:tan θ/tan(90°- θ)+sin(90°- θ)/cosθ=sec^2 θ​

Answer 1

Hey! Your answer refers to attachment :)

Answer 2

$\boxed{ \bold{ \huge{ \sf{ \orange{Answer}}}}} \\ \\ \star \sf \: To \: Prove \\ \\ \implies \sf \: \frac{ \tan \theta }{ \tan(90 - \theta) } + \frac{ \sin(90 - \theta)}{ \cos\theta } = { \sec }^{2} { \theta} \\ \\ \star \sf \: Formula \\ \\ \implies \sf \: \tan(90 - \theta) = \cot\theta \\ \\ \implies \sf \: \sin(90 - \theta) = \cos \theta \\ \\ \implies \sf \: \frac{ \tan \theta}{ \cot \theta } = { \tan }^{2} { \theta} \\ \\ \star \sf \: Step \: by \: Step \: Explanation \\ \\ \implies \sf \: \frac{ \tan \theta }{ \cot \theta} + \frac{ \cos \theta}{ \cos \theta} = { \tan }^{2} { \theta} + 1 \\ \\ \implies \sf \: { \tan }^{2}{ \theta} + 1 = { \sec}^{2} { \theta} \\ \\ \therefore \: \sf \: LHS = RHS$