Advertisements
Advertisements
प्रश्न
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
Advertisements
उत्तर
L.H.S. = `sqrt(sec^2A + cosec^2A)`
= `sqrt(1/cos^2A + 1/sin^2A)`
= `sqrt((sin^2A + cos^2A)/(sin^2Acos^2A)`
= `sqrt(1/(sin^2Acos^2A)`
= `sqrt(1/(sinAcosA))`
R.H.S. = tan A + cot A
= `sinA/cosA + cosA/sinA`
= `(sin^2A + cos^2A)/(sinAcosA)`
= `1/(sinAcosA)`
L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Evaluate:
`(tan 65°)/(cot 25°)`
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
