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Question
Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ
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Solution
L.H.S = \[\cot^2 \theta - \tan^2 \theta\]
[1 + tan2θ = sec2θ, 1 + cot2θ = coses2θ]
\[ = \left( {cosec}^2 \theta - 1 \right) - \left( \sec^2 \theta - 1 \right)\]
\[ = {cosec}^2 \theta - 1 - \sec^2 \theta + 1\]
\[ = {cosec}^2 \theta - \sec^2 \theta\]
= R.H.S
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Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
