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 that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
Without using trigonometric table , evaluate :
`cosec49°cos41° + (tan31°)/(cot59°)`
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
