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प्रश्न
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
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उत्तर
LHS = sec2 θ + cosec2 θ
= `1/(cos^2 θ) + 1/(sin^2 θ)`
= `(sin^2 θ + cos^2 θ)/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ) xx 1/(cos^2 θ)`
= sec2 θ cosec2 θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
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Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
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`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
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Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
