The numerical value of 1 + $$\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}}$$  - $${\text{se}}{{\text{c}}^2}{27^ \circ }$$  + $$\frac{1}{{{{\sin }^2}{{63}^ \circ }}}$$  - $${\text{cose}}{{\text{c}}^2}{27^ \circ }$$   is?

Solution (By Examveda Team)

$$\eqalign{ & {\text{1 + }}\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}} - {\text{se}}{{\text{c}}^2}{27^ \circ }{\text{ + }}\frac{1}{{{{\sin }^2}{{63}^ \circ }}} - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}{63^ \circ } - {\text{se}}{{\text{c}}^2}{27^ \circ } + {\text{cose}}{{\text{c}}^2}{63^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\sec ^2}{27^ \circ } + {\text{se}}{{\text{c}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$