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प्रश्न
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
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उत्तर
LHS = `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ))`
= `(sin θ. sin θ cos θ)/(cos θ) + (cos θ . cos θ sin θ)/(sin θ)`
= sin2 θ + cos2 θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Find A if tan 2A = cot (A-24°).
Prove that `(1 + sec A)/(sec A) = (sin^2A)/(1 - cos A)`.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
