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प्रश्न
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
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उत्तर
`(p^2 - 1)/(p^2 + 1)`
= `((secA + tanA)^2 - 1)/((secA + tanA)^2 + 1)`
= `(sec^2A + tan^2A + 2tanA secA - 1)/(sec^2A + tan^2A + 2tanA secA + 1)`
= `(tan^2A + tan^2A + 2tanA secA)/(sec^2A + sec^2A + 2tanA secA)`
= `(2tan^2A + 2tanA secA)/(2sec^2A + 2tanA secA)`
= `(2tanA(tanA + secA))/(2secA(tanA + secA)`
= sin A
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संबंधित प्रश्न
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
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Prove the following identities:
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Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
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∴ L.H.S. = R.H.S.
