Question

Evaluate: $\frac{({\sin }^{2}63°+{\sin }^{2}27°)}{({\cos }^{2}17°+{\cos }^{2}73°)}$

Answer 1

To evaluate:$\frac{({\sin }^{2}63°+{\sin }^{2}27°)}{({\cos }^{2}17°+{\cos }^{2}73°)}$

To simplify this trigonometric fraction, take complement of anyone trigonometric term from both numerator and denominator, we get

$\frac{({\sin }^{2}63°+{\sin }^{2}27°)}{({\cos }^{2}17°+{\cos }^{2}73°)}$ $=$$\frac{{\sin }^2(90°-{27}^0)+{\sin }^{2}27°}{{\cos }^2(90°-{73}^0)+{\cos }^{2}73°}$ (${63}^0$ can be written as ${90}^0-{27}^0$ & ${17}^0$ can be written as ${90}^0-{73}^0$)

We know that by complementary angle form, $\sin ({90}^0-\mathrm{\theta })=\mathrm{cos\theta }$ ⇒ ${\sin }^2({90}^0-\mathrm{\theta })={\cos }^2\mathrm{\theta }$ & $\cos ({90}^0-\mathrm{\theta })=\mathrm{sin\theta }$ ⇒${\cos }^2({90}^0-\mathrm{\theta })={\sin }^2\mathrm{\theta }$

Now$\frac{({\sin }^{2}63°+{\sin }^{2}27°)}{({\cos }^{2}17°+{\cos }^{2}73°)}$$=$$\frac{{\cos }^2{27}^0+{\sin }^{2}27°}{{\cos }^2{73}^0+{\cos }^{2}73°}$ (complement of anyone trigonometric term is taken on both numerator and denominator)

$=$$\frac{1}{1}$ (By identity ${\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1$)

Thus, the required value of $\frac{({\sin }^{2}63°+{\sin }^{2}27°)}{({\cos }^{2}17°+{\cos }^{2}73°)}$ $=$ $1$