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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 identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
Prove that:
`cosA/(1 + sinA) = secA - tanA`
` tan^2 theta - 1/( cos^2 theta )=-1`
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
