Advertisements
Advertisements
प्रश्न
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Advertisements
उत्तर
tan2A − sin2A = tan2A · sin2A
LHS:
tan2A − sin2A
We know that:
`tan A = (sin A)/(cos A) => tan^2A = (sin^2 A)/(cos^2 A)`
`tan^2 A - sin^2A = (sin^2 A)/(cos^2 A) - sin^2 A`
`= (sin^2 A - sin^2 A cos^2 A)/(cos^2 A)`
Factor out sin2A:
= `(sin^2 A(1 - cos^2 A))/(cos^2 A)`
1 − cos2A = sin2A
= `(sin^2 A * sin^2 A)/(cos^2 A) = (sin^4 A)/(cos^2 A)`
RHS:
tan2A − sin2A = tan2A · sin2A
APPEARS IN
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 + 1/tan theta` = sec θ.cosec θ
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
