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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 trigonometric identities.
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta)/cos theta`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
If 5 `tan theta = 4,"write the value of" ((cos theta - sintheta))/(( cos theta + sin theta))`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
If 3 sin θ = 4 cos θ, then sec θ = ?
Prove that `(sintheta + tantheta)/cos theta` = tan θ(1 + sec θ)
