2(\sin^{6}\theta+\cos^{6}\theta)-3(\sin^{4}\theta+\cos^{4}\theta)=

AP ICET 2015 Paper

Question : 121 of 200

Marks: +1, -0

2(\sin^{6}θ+\cos^{6}θ)-3(\sin^{4}θ+\cos^{4}θ)=

Solution:

2(\sin^{6}θ+\cos^{6}θ)-3(\sin^{4}θ+\cos^{4}θ)=

Taking

\sin^2θ=a,\cos^2θ=b

⇒a+b=1

=2(a^3+b^3)–3(a^2+b^2)

=2[(a+b)(a^{2}+b^{2}-2ab)]-3(a^{2}+b^{2})

=2(a^{2}+b^{2}-2ab)-3(a^{2}+b^{2})

[a+b=1]

=-(a^{2}+b^{2}+2ab)

=-(a+b)^{2}=-(1)^{2}=-1

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