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प्रश्न
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
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उत्तर
LHS = `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ`
= `(sin^4θ + cos^4θ)/(sin^2θ.cos^2θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2(sin^2 θ. cos^2 θ))/(sin^2 θ.cos^2 θ)`
= `((1)^2 - 2sin^2θ. cos^2 θ)/(sin^2 θ.cos^2 θ)`
= `1/(sin^2 θ.cos^2 θ) - (2sin^2θ. cos^2 θ)/(sin^2 θ.cos^2 θ)`
= `1/(sin^2 θ.cos^2 θ) - 2`
= RHS
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
Prove the following identities:
`(1 + cos theta - sin^2 theta )/(sin theta (1 + cos theta)) = cot theta`
If `m = (cos θ - sin θ)` and `n = (cos θ + sin θ)`, show that `sqrt(m/n) + sqrt(n/m) = 2/sqrt(1 - tan^2θ)`.
Write the value of tan1° tan 2° ........ tan 89° .
Prove the following identity :
`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
