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प्रश्न
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
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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.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Prove the following identity :
`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
