Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Advertisements
उत्तर
`(1 - cos^2θ)sec^2θ = tan^2θ`
Consider L.H.S = `sin^2θ1/cos^2θ`
= `tan^2θ` = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove that `sqrt(sec^2 theta + cosec^2 theta) = tan theta + cot theta`
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
If `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`, prove that `(m + n)^(2/3) + (m - n)^(2/3) = 2a^(2/3)`
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
