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
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Prove that sec2θ − cos2θ = tan2θ + sin2θ
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
