Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
tan2 θ − sin2 θ = tan2 θ sin2 θ
Prove that:
tan2 θ − sin2 θ = tan2 θ sin2 θ
Advertisements
उत्तर
LHS = tan2 θ − sin2 θ
= `sin^2 θ/cos^2 θ - sin^2 θ` `[∵ tan^2 θ = sin^2 θ/cos^2 θ]`
`=> sin^2 θ [1/cos^2 θ- 1]`
`sin^2 θ [(1 - cos^2 θ)/cos^2 θ]`
`=> sin^2 θ. sin^2 θ/cos^2 θ = sin^2 θ tan^2 θ `
LHS = RHS
Hence proved
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities:
(1 + tan2 θ) (1 − sin θ) (1 + sin θ) = 1
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
