Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
Advertisements
उत्तर
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
`(cotA + tanB)/(cotB + tanA)`
= `(1/tanA + tanB)/(1/tanB + tanA)`
= `((1 + tanAtanB)/tanA)/((1 + tanAtanB)/tanB) = (1 + tanAtanB)/tanA.tanB/(1 + tanAtanB)`
= `tanB/tanA = 1/tanA.tanB = cotAtanB`
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that (m2 + n2) = (a2 + b2).
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
Eliminate θ if x = r cosθ and y = r sinθ.
