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प्रश्न
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
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उत्तर
LHS = `sec^2A + cosec^2A`
= `1/cos^2A + 1/sin^2A = (sin^2A + cos^2A)/(cos^2A.sin^2A)`
= `1/(cos^2A.sin^2A) = sec^2Acosec^2A` = RHS
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संबंधित प्रश्न
`(1-cos^2theta) sec^2 theta = tan^2 theta`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
If 3 sin θ = 4 cos θ, then sec θ = ?
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
