Question

Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`

Answer 1

L.H.S = `sqrt(sec^2 theta + "cosec"^2 theta)`

= `sqrt(1/cos^2 theta + 1/(sin^2 theta))`  ...`[∵ sec^2 theta = 1/(cos^2 theta) "and"  "cosec"^2 theta = 1/(sin^2 theta)]`

= `sqrt((sin^2 theta + cos^2 theta)/(sin^2 theta * cos^2 theta))`

= `sqrt(1/(sin^2 theta * cos^2 theta))`  ...[∵ sin²θ + cos²θ = 1]

= `1/(sin theta * cos theta)`

= `(sin^2 theta + cos^2 theta)/(sin theta * cos theta)`  ...[∵ 1 = sin²θ + cos²θ]

= `(sin^2 theta)/(sin theta * cos theta) + (cos^2 theta)/(sin theta * cos theta)`

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

= tan θ + cot θ 

= R.H.S

Hence proved.