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प्रश्न
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
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उत्तर
LHS = `cosA/(1 - tanA) + sinA/(1 - cotA)`
= `cosA/(1-sinA/cosA) + sinA/(1 - cosA/sinA) = cosA/((cosA -sinA)/cosA) + sinA/((sinA - cosA)/sinA)`
= `cos^2A/(cosA - sinA) + sin^2A/(sinA - cosA) = (cos^2A - sin^2A)/((cosA - sinA))`
`((cosA - sinA)(cosA + sinA))/(cosA - sinA)`
= sinA + cosA = RHS
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संबंधित प्रश्न
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
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`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
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`cot^2A/(cosecA - 1) - 1 = cosecA`
If (tan θ + sin θ) = m and (tan θ – sin θ) = n, prove that (m2 – n2)2 = 16 mn.
What is the value of 9cot2 θ − 9cosec2 θ?
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
