Advertisements
Advertisements
प्रश्न
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Advertisements
उत्तर
LHS = sin (90° - θ) cos (90° - θ)
LHS = cos θ. sin θ
RHS = tan θ. cos2θ
RHS = `sin θ/cos θ` x cos2θ
RHS = cos θ. sin θ
∴ LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
If cosθ = `5/13`, then find sinθ.
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
