Advertisements
Advertisements
Question
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Advertisements
Solution
LHS = `(secA - 1)/(secA + 1)`
= `(1/cosA - 1)/(1/cosA + 1) = (1 - cosA)/(1 + cosA)`
= `(1 - cosA)/(1 + cosA) xx (1 + cosA)/(1 + cosA)`
= `(1-cos^2A)/(1 + cosA)^2`
= `sin^2A/(1 + cosA)^2` (∵ `1 - cos^2A = sin^2A`)
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Eliminate θ if x = r cosθ and y = r sinθ.
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
