Question

Evaluate : $\frac{cose{c}^2{63}^{\circ }+\tan {}^2{24}^{\circ }}{\cot {}^2{66}^{\circ }+\sec {}^2{27}^{\circ }}+\frac{\sin {}^2{63}^{\circ }+\cos {63}^{\circ }\sin {27}^{\circ }+\sin {27}^{\circ }\sec {63}^{\circ }}{2(cose{c}^2{65}^{\circ }-\tan {}^2{25}^{\circ })}$

Answer 1

$\frac{cose{c}^2{63}^{\circ }+ta{n}^2{24}^{\circ }}{co{t}^2{66}^{\circ }+se{c}^2{27}^{\circ }}+\frac{si{n}^2{63}^{\circ }+cos{63}^{\circ }sin{27}^{\circ }+sin{27}^{\circ }sec{63}^{\circ }}{2(cose{c}^2{65}^{\circ }-ta{n}^2{25}^{\circ })}$

$\Rightarrow \frac{cose{c}^2{63}^{\circ }+ta{n}^2{24}^{\circ }}{ta{n}^2(90-{66}^{\circ })+cose{c}^2({90}^{\circ }-{27}^{\circ })}+\frac{si{n}^2{63}^{\circ }+cos63cos(90-27)+sin{27}^{\circ }cosec(90-{63}^{\circ })}{2[cose{c}^2{65}^{\circ }-co{t}^2(90-25\left)\right]}$

$\Rightarrow$

We know tan(90-A)= cot A, cosec(90-A)= sec A & cot (90-A) = tan A

and $cose{c}^2\theta -co{t}^2\theta =1$

then

$\Rightarrow \frac{cose{c}^{2}63+ta{n}^{2}24}{ta{n}^{2}24+cose{c}^{2}63}+\frac{si{n}^{2}63+co{s}^2{63}^{\circ }+1}{2(cose{c}^{2}65-co{t}^2{65}^{\circ })}$

$\Rightarrow 1+\frac{2}{2\times 1}$

$\Rightarrow 2$