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प्रश्न
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
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उत्तर
Given : sin θ + cos θ = m and secθ + cosecθ = n
Consider L.H.S. = n(m2 - 1) = (secθ + cosecθ)[(sinθ + cosθ)2 - 1]
= `(1/cosθ + 1/sinθ) [sin^2θ + cos^2θ + 2sinθcosθ - 1`]
= `((cosθ + sinθ)/(sinθcosθ)) (1 + 2sinθcosθ - 1)`
= `((cosθ + sinθ))/(sinθcosθ) (2 sinθ cosθ)`
= 2(sinθ + cosθ)
= 2m = R.H.S.
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संबंधित प्रश्न
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
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.
