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प्रश्न
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
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उत्तर
LHS = `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A)`
= `((sin A + cos A)^2 + (sin A - cos A)^2)/((sin A - cos A)(sin A + cos A))`
= `(sin^2 A + cos^2 A + 2 sin A cos A + sin^2 A + cos^2 A - 2sin A. cos A)/(sin^2 A - cos^2 A)`
= `2(sin^2 A + cos^2 A)/(sin^A - cos^2 A)`
= `2/(sin^2 A - cos^2 A)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
`(sec^2 theta-1) cot ^2 theta=1`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
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.
