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प्रश्न
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
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उत्तर
sin3θ + cos3θ
= (sin θ + cos θ)(sin2θ + cos2 – sin θ cos θ)
= (sin θ + cos θ)(1 – sin θ cos θ). ...(i)
L.H.S = `(sin^3θ + cos^3θ)/(sinθ + cosθ) + sinθcosθ`
= `((sinθ + cosθ)(1 - sinθcosθ))/((sinθ + cosθ)) + sinθcosθ` ...(From(i))
= 1 – sin θ cos θ + sin θ.cos θ
Simplify by cancelling – sin θ cos θ and + sin θ.cos θ
= 1
= R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
