Question

Find the value of $\frac{co{s}^2{26}^{\circ }+co{s}^2{64}^{\circ }}{si{n}^2{55}^{\circ }+si{n}^2{35}^{\circ }}.$

• A. 1

Answer 1

The correct option is A 1
$\frac{co{s}^2{26}^{\circ }+co{s}^2{64}^{\circ }}{si{n}^2{55}^{\circ }+si{n}^2{35}^{\circ }}$

$\because$ $sin({90}^{\circ }-\theta )$ = $cos\left(\theta \right)$;
$sin({90}^{\circ }-\theta )$ = $cos\left(\theta \right)$

$=\frac{co{s}^2{26}^{\circ }+si{n}^2({90}^{\circ }-{64}^{\circ })}{co{s}^2({90}^{\circ }-{55}^{\circ })+si{n}^2{35}^{\circ }}$

$=\frac{co{s}^2{26}^{\circ }+si{n}^2{26}^{\circ }}{co{s}^2{35}^{\circ }+si{n}^2{35}^{\circ }}$

By Trigonometric Identity,

[ $si{n}^2\theta +co{s}^2\theta =1$]

$=\frac{1}{1}$

$=1$