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Question
Prove that:
Solution
\[\cot\theta + \tan\theta\]
\[ = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\]
\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\sin\theta\cos\theta}\]
\[ = \frac{1}{\sin\theta\cos\theta} \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]
\[ = \frac{1}{\sin\theta} \times \frac{1}{\cos\theta}\]
\[ = \text{ cosec } \theta\sec\theta\]
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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))`
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= `square/(square square)`
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= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ