Advertisements
Advertisements
प्रश्न
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Advertisements
उत्तर
LHS = `sqrt(2 + tan^2 θ + cot^2 θ)`
= `sqrt( tan^2 θ + cot^2θ + 2tan θ.cot θ)` ...[ ∵ tan θ.cot θ = 1 ]
= `sqrt( tan^2 θ + cot^2θ)`
= tan θ + cot θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + tan theta + cot theta`
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
tan2 A + cot2 A = sec2 A cosec2 A − 2
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Define an identity.
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
