Advertisements
Advertisements
प्रश्न
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
उत्तर
LHS = 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1
= 2( sin2θ + cos2θ ) [ sin4θ + cos4θ - sin2θ.cos2θ ] - 3[ ( sin2θ + cos2θ )2 - 2sin2θ. cos2θ + 1
= 2 x 1 [ ( sin2θ + cos2θ )2 - 2 sin2θ.cos2θ - sin2θ.cos2θ ] - 3[ (1)2 - 2sin2θ. cos2θ ] + 1
= 2 [ (1)2 - 3 sin2θ.cos2θ ] - 3 [ 1 - 2 sin2θ. cos2θ ] + 1
= 2 - 6 sin2θ. cos2θ - 3 + 6 sin2θ. cos2θ + 1
= - 1 + 1 = 0
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
What is the value of (1 − cos2 θ) cosec2 θ?
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2
The value of sin2 29° + sin2 61° is
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
Show that tan4θ + tan2θ = sec4θ – sec2θ.
