Advertisements
Advertisements
Question
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Solution
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
RELATED QUESTIONS
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
` tan^2 theta - 1/( cos^2 theta )=-1`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
If `sec theta + tan theta = x," find the value of " sec theta`
Without using trigonometric table , evaluate :
`(sin49^circ/sin41^circ)^2 + (cos41^circ/sin49^circ)^2`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
