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प्रश्न
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
उत्तर
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
LHS = `(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1)`
= `(cosec^2A + cosecA + cosec^2A - cosecA)/(cosec^2A - 1)`
= `(2cosec^2A)/cot^2A(Q cosec^2A - 1 = cot^2A)`
= `(2/sin^2A)/(cos^2A/sin^2A) = 2/cos^2A = 2sec^2A`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
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Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
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If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
