Trigonometry – II — Maths STD 11 Science — Question

\begin{aligned}
& (\sin x-\cos x)^4 \\
& =\left[(\sin x-\cos x)^2\right]^2 \\
& =\left(\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right)^2 \\
& =(1-2 \sin x \cos x)^2 \\
& =1-4 \sin x \cos x+4 \sin ^2 x \cos ^2 x \\
& (\sin x+\cos x)^2=\sin ^2 x+\cos ^2 x+2 \sin x \cos x=1+2 \sin x \cos x \\
& \sin ^6 x+\cos ^6 x \\
& =\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\
& =\left(\sin { }^2 x+\cos ^2 x\right)^3-3 \sin ^2 x \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right) \ldots . . \because a^3+ \\
& \left.b^3=(a+b)^3-3 a b(a+b)\right] \\
& =1^3-3 \sin ^2 x \cos ^2 x(1) \\
& =1-3 \sin ^2 x \cos ^2 x \\
& \text { L.H.S. }=3\left(\sin ^2 x-\cos x\right)^4+6(\sin x+\cos x)^2+4\left(\sin { }^6 x+\cos ^6 x\right) \\
& =3\left(1-4 \sin ^2 x \cos ^2 x+4 \sin ^2 x \cos ^2 x\right)+6\left(1+2 \sin ^2 x \cos x\right)+ \\
& 4\left(1-3 \sin ^2 x \cos ^2 x\right) \\
& =3-12 \sin ^2 x \cos ^2 x+12 \sin ^2 x \cos ^2 x+6+12 \sin ^2 x \cos x+4- \\
& 12 \sin ^2 x \cos ^2 x \\
& =13 \\
& =\text { R.H.S. }
\end{aligned}