Advertisements
Advertisements
प्रश्न
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Advertisements
उत्तर
LHS = `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2`
⇒ `(1 + sin^2 θ + cos^2 θ + 2(sin θ - cos θ - sin θ. cos θ))/(1 + sin^2 θ + cos^2 θ + 2(sin θ + cos θ + sin θ. cos θ)`
= `(1 + 1 + 2 (sin θ - cos θ - sin θ. cos θ))/( 1 + 1 + 2((sin θ + cos θ + sin θ. cos θ)`
= `(2 (1 + sin θ - cos θ - sin θ. cos θ))/(2( 1 + (sin θ + cos θ + sin θ. cos θ))`
= `( 1 + sin θ - cos θ( 1 + sin θ))/(1 + sin θ + cos θ( 1 + sin θ))`
= `((1 + sin θ)(1 - cos θ))/((1 + sin θ)( 1 + cos θ))`
= `(1 - cos θ)/( 1 + cos θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following identities:
`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`
`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`
`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Choose the correct alternative:
sec 60° = ?
If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
