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प्रश्न
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
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उत्तर
LHS = sin θ sin( 90° - θ) - cos θ cos( 90° - θ)
= sin θ . cos θ - cos θ . sin θ
= 0
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
