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प्रश्न
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
Prove the following identities:
`(sin A)/((sec A + tan A - 1)) + (cos A)/(("cosec" A + cot A - 1)) = 1`
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उत्तर
LHS = `(sec A)/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1)`
= `(sin A)/(1/cos A + sin A/cos A - 1) + cos A/(1/sin A + cos A/sin A - 1)`
= `(sin A/(1 + sin A - cos A))/cos A + ((cos A)/(1 + cos A - sin A))/(sin A)`
= `(sin A.cos A)/(1 + sin A - cos A) + (sin A. cos A)/(1 + cos A - sin A)`
= `(sin A. cos A( 1 + cos A - sin A + 1 + sin A - cos A))/([ 1 + (sin A - cos A)][1 - (sin A - cos A)])`
= `(2sin A. cos A)/((1)^2 - (sin A - cos A)^2)`
= `(2sin A. cos A)/(1 - (sin^2 A + cos^2 A - 2 sin A.cos A))`
= `(2 sin A. cos A)/(1 - 1 + 2 sin A. cos A)`
= `2/2 = 1`
= RHS
Hence proved.
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संबंधित प्रश्न
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
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`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
The value of sin2θ + `1/(1 + tan^2 theta)` is equal to
tan θ cosec2 θ – tan θ is equal to
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)
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
