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प्रश्न
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
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उत्तर
\[\left( \sec\theta + \tan\theta \right)\left( 1 - \sin\theta \right)\]
\[ = \left( \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \right)\left( 1 - \sin\theta \right)\]
\[ = \left( \frac{1 + \sin\theta}{\cos\theta} \right)\left( 1 - \sin\theta \right)\]
\[ = \frac{1 - \sin^2 \theta}{\cos\theta}\]
\[ = \frac{\cos^2 \theta}{\cos\theta} \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]
\[ = \cos\theta\]
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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)`
= 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
