Advertisements
Advertisements
प्रश्न
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
उत्तर
LHS = sec2 θ + cosec2 θ
= `1/(cos^2 θ) + 1/(sin^2 θ)`
= `(sin^2 θ + cos^2 θ)/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ) xx 1/(cos^2 θ)`
= sec2 θ cosec2 θ
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`(cos^3 theta +sin^3 theta)/(cos theta + sin theta) + (cos ^3 theta - sin^3 theta)/(cos theta - sin theta) = 2`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
What is the value of (1 + cot2 θ) sin2 θ?
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2