Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Advertisements
उत्तर
LHS = `(1 + tan^2θ)sinθcosθ`
= `(1 + sin^2θ/cos^2θ)sinθcosθ`
= `((cos^2θ + sin^2θ)/cos^2θ)sinθcosθ`
= `1/cos^2θ xx sinθcosθ` (∵ `cos^2θ + sin2θ = 1`)
= `sinθ/cosθ = tanθ`
APPEARS IN
संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove that:
`cosA/(1 + sinA) = secA - tanA`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
\[\frac{x^2 - 1}{2x}\] is equal to
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
If 2 cos θ + sin θ = `1(θ ≠ π/2)`, then 7 cos θ + 6 sin θ is equal to ______.
