Advertisements
Advertisements
प्रश्न
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
Advertisements
उत्तर
sin θ (1 + sin2 θ) = cos2 θ
sin θ (1 + 1 – cos2 θ) = cos2 θ
sin θ (2 – cos2 θ) = cos2 θ
Squaring on both sides,
sin2 θ (2 – cos2 θ)2 = cos4 θ
(1 – cos2 θ) (4 + cos4 θ – 4 cos2 θ) = cos4 θ
4 cos4 θ – 4 cos2 θ – cos6 θ + 4 cos4 θ = cos4 θ
4 + 5 cos4 θ – 8 cos2 θ – cos6 θ = cos4 θ
– cos6 θ + 5cos4 θ – cos4 θ – 8 cos2 θ = – 4
– cos6 θ + 4cos4 θ – 8 cos2 θ = – 4
cos6 θ – 4cos4 θ + 8 cos2 θ = 4
Hence it is proved
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities:
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
Prove the following identities:
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta )/(cos theta)`
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
