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प्रश्न
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
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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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संबंधित प्रश्न
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`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
