\(\text { We know that in } \Delta \mathrm{ABC}\)
\(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}>0\)
\(\text { And also } \sin A \sin B \sin C>0\)
Both are true..
Since sine is positive in I and II quadrant.
Both (i) and (ii) are true.

In one second it rotates = 2 radians.
For 2 radians it takes 1 second.
\(\text { For } 2 \pi \text { ( } 1 \text { revolution) it will take } \frac{2 \pi}{2}=\pi \text { second. }\)
\(\therefore \text { For } 10 \text { revolutions it takes } 10 \pi \text { seconds. }\)

\(\mathrm{f}_{4}(x)-\mathrm{f}_{6}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]-\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right] \)
\(=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right]-\frac{1}{6}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{3}\right. \)
\(\left.=-3 \sin ^{2} x \cos ^{2} x+\left(\sin ^{2} x+\cos ^{2} x\right)\right] \)
\(=\frac{1}{4}\left(1-2 \sin ^{2} x \cos ^{2} x\right)-\frac{1}{6}\left(1-3 \sin ^{2} x \cos ^{2} x\right) \)
\(=\frac{1}{4}-\frac{1}{2} \sin ^{2} x \cos ^{2} x-\frac{1}{6}+\frac{1}{2} \sin ^{2} x \cos ^{2} x \)
\(=\frac{1}{4}-\frac{1}{6} \)
\(=\frac{3-2}{12}=\frac{1}{12} \)

\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} =\tan \left(140^{\circ}-130^{\circ}\right) \)
\(=\tan 10^{\circ} \ldots \ldots(1) \)
\(\tan 40^{\circ} =\lambda \)
\(\tan 80^{\circ}=\frac{2 \tan 40^{\circ}}{1-\tan ^{2} 40^{\circ}} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\tan \left(90^{\circ}-10^{\circ}\right) =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\cot 10^{\circ} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\Rightarrow \tan 10^{\circ} =\frac{1-\lambda^{2}}{2 \lambda} \)
\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} \)
\(=\frac{1-\lambda^{2}}{2 \lambda}\)

\(\cos 28^{\circ}+\sin 28^{\circ} =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\sin \left(90^{\circ}-62^{\circ}\right) =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\cos 62^{\circ} =\mathrm{k}^{3} \)
\(2 \cos 45^{\circ} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(2 \frac{1}{\sqrt{2}} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(\cos 17^{\circ} =\frac{\mathrm{k}^{3}}{\sqrt{2}} \)