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प्रश्न
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
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उत्तर
LHS = `(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))`
=`tan^2 theta/sec^2 theta + cot^2 theta/ cosec ^2 theta (∵ sec^2 theta - tan^2 theta = 1 and cosec^2 theta - cot^2 theta=1)`
=`(sin^2theta/cos^2 theta)/(1/cos^2theta) + (cos^2theta/sin^2 theta)/(1/sin^2 theta)`
=` sin^2 theta + cos^2 theta`
=1
= RHS
Hence, LHS = RHS
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संबंधित प्रश्न
Prove the following trigonometric identities:
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Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
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