Question

Prove the following:

tan θ + tan (90° – θ) = sec θ sec (90° – θ)

Answer 1

L.H.S = tan θ + tan (90° – θ)  ...[∵ tan (90° – θ) = cot θ]

= tan θ + cot θ

= `sinθ/cosθ + cosθ/sinθ`

= `(sin^2 theta + cos^2 theta)/(sin theta cos theta)`  ...`[∵ tan theta = sintheta/costheta "and" cot theta = costheta/sintheta]`

= `1/(sin theta cos theta)`   ...[∵ sin²θ + cos²θ = 1]

= sec θ cosec θ   ...`[∵ sec theta = 1/costheta "and" cos theta = 1/sin theta]`

= sec θ sec (90° – θ)  ...[∵ sec (90° – θ) = cosec θ] 

= R.H.S