Advertisements
Advertisements
प्रश्न
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Advertisements
उत्तर
LHS = cosec2(90° - θ) - tan2 θ
LHS = sec2 θ - tan2 θ = 1
RHS = cos2(90° - θ) + cos2 θ
RHS = sin2θ + cos2 θ = 1
Hence, LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
