Advertisements
Advertisements
प्रश्न
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Advertisements
उत्तर
LHS = `1/(sin θ + cos θ) + 1/(sin θ - cos θ)`
= `((sin θ - cos θ) + (sin θ + cos θ))/(sin^2 θ - cos^2 θ)`
= `(2 sin θ)/((1 - cos^2 θ) - cos^2 θ)`
= `(2 sin θ)/(1 - 2cos^2 θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
