If $$\frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}{{20}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}{{31}^ \circ }} \right)}}$$     = $$\frac{2}{m}{\text{,}}$$  then m is equal to?

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}{{20}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}{{31}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}\left( {{{90}^ \circ } - {{59}^ \circ }} \right)} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{ta}}{{\text{n}}^2}{{70}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\cot }^2}{{59}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{1}{2} = \frac{2}{m}\left[ {{{\sec }^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1} \right] \cr & (cose{c^2}\theta - {\cot ^2}\theta = 1) \cr & \Rightarrow m = 2 \times 2 \cr & \Rightarrow m = 4 \cr} $$