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प्रश्न
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
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उत्तर
LHS = `(1 - tan^2 θ)/(cot^2 θ - 1)`
= `(1 - tan^2 θ)/(1/tan^2 θ - 1)`
= `((1 - tan^2 θ)/(1 - tan^2 θ)/tan^2 θ) `
= tan2 θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
Prove the following identities:
`(cos theta "cosec" theta - sin theta sec theta )/(cos theta + sin theta) = "cosec" theta - sec theta`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
