Advertisements
Advertisements
प्रश्न
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Advertisements
उत्तर
LHS = `(sin B + sec A)/sin A`
= `(sin (90 - A) + sec A)/sin A`
= `(cos A + sec A)/sin A`
= `(cos^2 A + 1)/(sin A. cos A)`
= `(2cos^2 A + sin^2 A)/(sin A. cos A)`
= 2cot A + tan A
= 2 tan B + tan A = RHS
hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities.
if cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
cos θ . sec θ = ?
Prove that `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`.
Prove that sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A.
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
