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प्रश्न
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
उत्तर
L.H.S. = sec2 A . cosec2 A
= `1/(cos^2A) * 1/(sin^2A)`
= `1/(cos^2A sin^2A)`
= `(sin^2A + cos^2A)/(cos^2A sin^2A)`
= `1/(cos^2A) + 1/(sin^2A)`
= sec2 A + cosec2 A
= 1 + tan2 A + 1 + cot2 A ...(∵ sec2 A = 1 + tan2 A and cosec2 A = 1 + cot2 A)
= tan2 A + cot2 A + 2 = R.H.S.
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संबंधित प्रश्न
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If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
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
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
