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 trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.
