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प्रश्न
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
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उत्तर
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.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
