Advertisements
Advertisements
प्रश्न
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
उत्तर
`1/"a" = cos theta/(1 + sin theta)`
Squaring on both sides,
`1/"a"^2 = (cos^2theta)/(1 + sin theta)^2= (1 - sin^2theta)/(1 + sin theta)^2`
`1/"a"^2 = ((1 + sin theta)(1 - sin theta))/(1 + sin theta)^2 = ((1 - sin theta))/((1 + sin theta))`
a2(1 − sin θ) = 1 + sin θ
⇒ a2 = `((1 + sin theta))/((1 - sin theta))`
L.H.S = `("a"^2 - 1)/("a"^2 + 1)`
= `((1 + sin theta))/((1 - sin theta)) - 1 ÷ ((1 + sin theta))/((1 - sin theta)) + 1`
= `((1 + sin theta) - (1 - sin theta))/((1 - sin theta)) ÷ ((1 + sin theta) + (1 - sin theta))/((1 - sin theta))`
= `(1 + sin theta - 1 + sin theta)/((1 - sin theta)) ÷ (1 + sin theta + 1 - sin theta)/((1 - sin theta))`
= `(2 sin theta)/(1 - sin theta) ÷ 2/(1 - sin theta)`
= `(2 sin theta)/(1 - sin theta) xx (1 - sin theta)/2`
= sin θ
∴ `("a"^2 - 1)/("a"^2 + 1)` = sin θ.
Hence it is proved.
APPEARS IN
संबंधित प्रश्न
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`
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
cos4 A − sin4 A is equal to ______.
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
(1 + sin A)(1 – sin A) is equal to ______.
