Advertisements
Advertisements
Question
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
Advertisements
Solution
`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
RELATED QUESTIONS
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`1/(tanA + cotA) = sinAcosA`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ
Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
