Question

$\frac{si{n}^2{63}^{\circ }+si{n}^2{27}^{\circ }}{co{s}^2{17}^{\circ }+co{s}^2{73}^{\circ }}$ =

• A. 0
• B. 1
• C. $\frac{1}{2}$

Answer 1

The correct option is B 1

We know that

$sin\theta =cos({90}^{\circ }-\theta )$, and

$cos\theta =sin({90}^{\circ }-\theta )$

Now the given expression is
$\frac{si{n}^2{63}^{\circ }+si{n}^2{27}^{\circ }}{co{s}^2{17}^{\circ }+co{s}^2{73}^{\circ }}$

$sin{27}^{\circ }=cos({90}^{\circ }-{27}^{\circ })\Rightarrow sin{27}^{\circ }=cos{63}^{\circ }$

and,
$cos{17}^{\circ }=sin({90}^{\circ }-{17}^{\circ })\Rightarrow cos{17}^{\circ }=sin{73}^{\circ }$

$\therefore \frac{si{n}^2{63}^{\circ }+si{n}^2{27}^{\circ }}{co{s}^2{17}^{\circ }+co{s}^2{73}^{\circ }}=\frac{si{n}^2{63}^{\circ }+co{s}^2{63}^{\circ }}{si{n}^2{73}^{\circ }+co{s}^2{73}^{\circ }}$

Now we know that $si{n}^2\theta +co{s}^2\theta =1$

$\therefore \frac{si{n}^2{63}^{\circ }+co{s}^2{63}^{\circ }}{si{n}^2{73}^{\circ }+co{s}^2{73}^{\circ }}=\frac{1}{1}=1$