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Question : 20 of 160
Marks:
+1,
-0
Solution:
2(sin6θ+cos6θ)−3(sin4θ+cos4θ) . . . (i)
a3+b3=(a+b)(a2+b2−ab)
(sin2θ)3+(cos2θ)3=(sin2θ+cos2θ)
(sin4θ+cos4θ−sin2θcos2θ)
sin6θ+cos6θ=sin4θ+cos4θ−sin2θcos2θ
Expression Eq. (i), 2sin4θ+2cos4θ
−2sin2θcos2θ−3sin4θ−3cos4θ
=−sin4θ−cos4θ−2sin2θcos2θ
=−(sin2θ+cos2θ)=−1
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