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Question
Prove that `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`.
Theorem
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Solution
R.H.S. = `(sec^2A)/("cosec"^2A)`
= `(1 + tan^2A)/(1 + cot^2A)` ...`[(∵ 1 + tan^2A = sec^2A),(1 + cot^2A = "cosec"^2A)]`
= `(1 + (sin^2A)/(cos^2A))/(1 + (cos^2A)/(sin^2A))`
= `((cos^2A + sin^2A)/(cos^2A))/((sin^2A + cos^2A)/(sin^2A))`
= `(1/(cos^2A))/(1/(sin^2A))` ...[∵ sin2A + cos2A = 1]
= `(sin^2A)/(cos^2A)`
= tan2A
= tan A . tan A
= `(tan A)/(cot A)`
= L.H.S.
∴ `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`
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