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Question
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
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Solution
LHS = `cosA/(1 - tanA) + sin^2A/(sinA - cosA)`
= `cosA/(1 - sinA/cosA) + sin^2A/(sinA - cosA)`
= `cosA/((cosA - sinA)/(cosA)) + sin^2A/(sinA - cosA)`
= `cos^2A/((cosA - sinA)) - sin^2A/((cosA - sinA))`
= `(cos^2A - sin^2A)/(cosA - sinA) = ((cosA + sinA)(cosA - sinA))/((cosA - sinA))`
= (cosA + sinA)
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RELATED QUESTIONS
Prove the following trigonometric identities.
`tan A/(1 + tan^2 A)^2 + cot A/((1 + cot^2 A)) = sin A cos A`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
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)
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 the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
