Advertisements
Advertisements
प्रश्न
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
उत्तर
L.H.S. `(1 - 2sin^2A)^2/(cos^4A - sin^4A)`
= `(1 - 2sin^2A)^2/((cos^2A)^2 - (sin^2A)^2`
= `(1 - 2sin^2A)^2/((cos^2A + sin^2A)(cos^2A - sin^2A))`
= `(1 - 2sin^2A)^2/((1)(1 - sin^2A - sin^2A)` ...[∵ cos2 A = 1 – sin2 A]
= `(1 - 2sin^2A)^2/((1 - 2sin^2A))`
= 1 – 2 sin2 A
= 1 – 2 (1 – cos2 A)
= 1 – 2 + 2 cos2 A
= 2 cos2 A – 1 = R.H.S.
Hence the result is proved.
APPEARS IN
संबंधित प्रश्न
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
Prove the following trigonometric identities.
`tan A/(1 + tan^2 A)^2 + cot A/((1 + cot^2 A)) = sin A cos A`
Write the value of tan10° tan 20° tan 70° tan 80° .
Simplify : 2 sin30 + 3 tan45.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
