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
संबंधित प्रश्न
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that (m2 + n2) = (a2 + b2).
Define an identity.
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
tan (90 – 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 `(cot A)/(1 - cot A) + (tan A)/(1 - tan A) = -1`.
