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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.
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