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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 identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identities:
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
Prove that:
`cosA/(1 + sinA) = secA - tanA`
Prove the following identities:
`(sin theta + 1 - cos theta)/(cos theta - 1 + sin theta) = (1 + sin theta)/(cos theta)`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
