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