Question

Without using trigonometric tables, evaluate that :
$\frac{{\sin }^2{20}^{\circ }+{\sin }^2{70}^{\circ }}{{\cos }^2{20}^{\circ }+{\cos }^2{70}^{\circ }}+\frac{\sin ({90}^{\circ }-\theta )\sin \theta }{\tan \theta }+\frac{\cos ({90}^{\circ }-\theta )\cos \theta }{\cot \theta }$

Answer 1

We have

$\frac{si{n}^2{20}^{\circ }+si{n}^2{70}^{\circ }}{co{s}^2{20}^{\circ }+co{s}^2{70}^{\circ }}+\frac{sin({90}^{\circ }-\theta )sin\theta }{tan\theta }+\frac{cos({90}^{\circ }-\theta )cos\theta }{cot\theta }$

$=\frac{si{n}^2{20}^{\circ }+si{n}^2({90}^{\circ }-{20}^{\circ })}{co{s}^2{20}^{\circ }+co{s}^2({90}^{\circ }-{20}^{\circ })}+\frac{sin({90}^{\circ }-\theta )sin\theta }{tan\theta }+\frac{cos({90}^{\circ }-\theta )cos\theta }{cot\theta }$

$=\frac{si{n}^2{20}^{\circ }+co{s}^2{20}^{\circ }}{co{s}^2{20}^{\circ }+si{n}^2{20}^{\circ }}+\frac{cos\theta sin\theta }{\frac{sin\theta }{cos\theta }}+\frac{sin\theta cos\theta }{\frac{cos\theta }{sin\theta }}$

$[\because sin({90}^{\circ }-\theta )=cos\theta andcos({90}^{\circ }-\theta )=sin\theta ]$

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