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प्रश्न
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
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उत्तर
L.H.S. = `sqrt((1 - cosA)/(1 + cosA))`
= `sqrt((1 - cosA)/(1 + cosA) xx (1 - cosA)/(1 - cosA))`
= `sqrt((1 - cosA)^2/(1 - cos^2A))`
= `sqrt((1 - cosA)^2/(sin^2A)`
= `(1 - cosA)/sinA`
= `1/sinA - cosA/sinA`
= cosec A – cot A = R.H.S.
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove the following trigonometric identities:
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
