Question

Evaluate : $\frac{{cosec}^2{63}^0+{tan}^2{24}^0}{{cot}^2{66}^0+{sec}^2{27}^0}+\frac{{sin}^2{63}^0+cos{63}^0{sin27}^0+{sin27}^0{sec63}^0}{2({cosec}^2{65}^0-{tan}^2{25}^0)}$.

Answer 1

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

$=\frac{se{c}^2({90}^o-{63}^o)+ta{n}^2{24}^o}{ta{n}^2({90}^o-{66}^o)+se{c}^2{27}^o}+\frac{si{n}^2{63}^o+cos{63}^{o}cos({90}^o-{27}^o)+sin{27}^{o}cosec({90}^o-{63}^o)}{2\left(se{c}^2\right({90}^o-{65}^o)-ta{n}^2{25}^o)}$

$=\frac{se{c}^2{27}^o+ta{n}^2{24}^o}{se{c}^2{27}^o+ta{n}^2{24}^o}+\frac{si{n}^2{63}^o+co{s}^2{63}^o+sin{27}^{o}cosec{27}^o}{2(se{c}^2{25}^o-ta{n}^2{25}^o)}$

$=1+\frac{1+1}{2}$

$=1+1=2$