Solution (By Examveda Team)
$$\eqalign{
& {\text{According to the question ,}} \cr
& {\text{ se}}{{\text{c}}^4}{\text{A}}\left( {1 - {{\sin }^4}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}} \cr
& {\text{Put A}} = {45^ \circ } \cr
& \Rightarrow {\text{ se}}{{\text{c}}^4}{45^ \circ }\left( {1 - {{\sin }^4}{{45}^ \circ }} \right) - 2{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr
& \Rightarrow 4\left( {1 - \frac{1}{4}} \right) - 2 \cr
& \Rightarrow 4 \times \frac{3}{4} - {\text{2}} \cr
& \Rightarrow 3 - 2 \cr
& \Rightarrow 1 \cr} $$
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