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प्रश्न
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
उत्तर
\[ = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\]
\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta} \times \frac{1}{\sin^2 \theta}\]
\[ = \sec^2 \theta \text{ cosec }^2 \theta\]
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संबंधित प्रश्न
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sin \[\theta\] cosec \[\theta\]= ?
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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
