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प्रश्न
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
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उत्तर
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
= `sin^2A + cosec^2A + 2sinA xx 1/sinA + cos^2A + sec^2A + 2cosA xx 1/cosA`
= sin2 A + cos2 A + cosec2 A + sec2 A + 2 + 2 ...(∵ sin2 A + cos2 A = 1)
= 1 + cosec2 A + sec2 A + 4
= (1 + cot2 A) + (1 + tan2 A) + 5 ...[∵ cosec2 A = 1 + cot2 A and sec2 A = 1 + tan2 A]
= 1 + cot2 A + 1 + tan2 A + 5
= 7 + tan2 A + cot2 A
= R.H.S.
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संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
` tan^2 theta - 1/( cos^2 theta )=-1`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
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
