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प्रश्न
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
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उत्तर
L.H.S. = `1/(sinA + cosA) + 1/(sinA - cosA)`
= `(sinA - cosA + sinA + cosA)/((sinA + cosA)(sinA - cosA))`
= `(2sinA)/(sin^2A - cos^2A)`
= `(2sinA)/(1 - cos^2A - cos^2A)` ...(∵ sin2A = 1 – cos2A)
= `(2sinA)/(1 - 2cos^2A)`
= R.H.S.
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संबंधित प्रश्न
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
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`cosecA + cotA = 1/(cosecA - cotA)`
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If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
