Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

The expression can be written as $E=\frac{\sin 50^{\circ }\sin 70^{\circ }+\sin 10^{\circ }\sin 70^{\circ }-\sin 10^{\circ }\sin 50^{\circ }}{\sin 10^{\circ }\sin 50^{\circ }\sin 70^{\circ }}$. First, calculate the denominator using the identity $\sin x\sin (60^{\circ }-x)\sin (60^{\circ }+x)=\frac{1}{4}\sin 3x$. For $x=10^{\circ }$, we have $\sin 10^{\circ }\sin 50^{\circ }\sin 70^{\circ }=\frac{1}{4}\sin (3\cdot 10^{\circ })=\frac{1}{4}\sin 30^{\circ }=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}$. Now, calculate the numerator $N$:$N=\sin 50^{\circ }\sin 70^{\circ }+\sin 10^{\circ }\sin 70^{\circ }-\sin 10^{\circ }\sin 50^{\circ }$. Using$2\sin A\sin B=\cos (A-B)-\cos (A+B)$:$2N=(\cos 20^{\circ }-\cos 120^{\circ })+(\cos 60^{\circ }-\cos 80^{\circ })-(\cos 40^{\circ }-\cos 60^{\circ })$$2N=\cos 20^{\circ }-(-1/2)+(1/2-\cos 80^{\circ })-(\cos 40^{\circ }-1/2)$$2N=\cos 20^{\circ }+1/2+1/2-\cos 80^{\circ }-\cos 40^{\circ }+1/2$$2N=\cos 20^{\circ }-\cos 80^{\circ }-\cos 40^{\circ }+3/2$. Using $\cos A-\cos B=2\sin ((A+B)/2)\sin ((B-A)/2)$:$\cos 20^{\circ }-\cos 80^{\circ }=2\sin 50^{\circ }\sin 30^{\circ }=2\sin 50^{\circ }(1/2)=\sin 50^{\circ }$. Since$\sin 50^{\circ }=\cos (90^{\circ }-50^{\circ })=\cos 40^{\circ }$.$2N=\sin 50^{\circ }-\cos 40^{\circ }+3/2=\cos 40^{\circ }-\cos 40^{\circ }+3/2=3/2$. So,$N=3/4$. Finally, $E=\frac{N}{\text{Denominator}}=\frac{3/4}{1/8}=\frac{3}{4}\cdot 8=6$.