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प्रश्न
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
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उत्तर
L.H.S. = (cosec A – sin A) (sec A – cos A) × sec2 A
= `(1/sinA - sinA)(1/cosA - cosA) xx sec^2A` ...`{∵ cosec theta = 1/sintheta, sectheta = 1/costheta, 1 - sin^2theta = cos^2theta, 1 - cos^2theta = sin^2theta}`
= `((1 - sin^2A)/sinA)((1 - cos^2A)/cosA) 1/(cos^2A)`
= `(cos^2A)/(sinA)*(sin^2A)/(cosA)*1/(cos^2A)`
= `sinA/cosA`
= tan A = R.H.S.
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संबंधित प्रश्न
Prove the following trigonometric identities:
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
Prove the following identities:
`(1 + cos theta - sin^2 theta )/(sin theta (1 + cos theta)) = cot theta`
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that (m2 + n2) = (a2 + b2).
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Find the value of sin 30° + cos 60°.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
