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प्रश्न
Prove the following.
उत्तर
\[\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta}\]
\[ = \frac{1 + \sin\theta + 1 - \sin\theta}{\left( 1 - \sin\theta \right)\left( 1 + \sin\theta \right)}\]
\[ = \frac{2}{1 - \sin^2 \theta} \left[ \left( a - b \right)\left( a + b \right) = a^2 - b^2 \right]\]
\[ = \frac{2}{\cos^2 \theta} \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]
\[ = 2 \sec^2 \theta\]
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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))`
= `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