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Question
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
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Solution
LHS = `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2`
= `tan A/(sec^2 A)^2 + cot A/(cosec^2 A)^2`
= `sin A/cos A xx cos^2 A xx cos^2 A + cos A/sin A xx sin^2 A xx sin^2 A`
= sin A.cos3A + sin3A.cos A
= sin A cos A (cos2 A + sin2 A)
= sin A. cos A x 1
= sin A. cos A
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Prove that:
`cosA/(1 + sinA) = secA - tanA`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
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
