Advertisements
Advertisements
प्रश्न
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Advertisements
उत्तर
LHS = `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ)`
= `(tan θ)/(tan θ) + ( cosec θ. cos θ)/(cosθ. cosec θ)`
= 1 + 1 = 2
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove the following trigonometric identities.
tan2 θ − sin2 θ = tan2 θ sin2 θ
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
cos4 A − sin4 A is equal to ______.
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
If cosA + cos2A = 1, then sin2A + sin4A = 1.
