Prove that sec 4 A ≤ ft 1-sin 4 A -2tan 2 A =1.

Consider the L.H.S,\(={{\sec }^{4}}A\left(1-{{\sin }^{4}}A \right)-2{{\tan }^{2}}A\)\( ={{\sec }^{4}}A-{{\sec}^{4}}A{{\sin }^{4}}A-2{{\tan }^{2}}A \)\( ={{\sec}^{4}}A-\dfrac{{{\sin }^{4}}A}{{{\cos }^{4}}A}-2{{\tan }^{2}}A \)\( ={{\sec }^{4}}A-{{\tan}^{4}}A-2{{\tan }^{2}}A \)\( =\left[ {{\left({{\sec }^{2}}A \right)}^{2}}-{{\left( {{\tan }^{2}}A \right)}^{2}}\right]-2{{\tan }^{2}}A \)\( =\left[ \left( {{\sec}^{2}}A \right)-\left( {{\tan }^{2}}A \right) \right]\left[ \left( {{\sec}^{2}}A \right)+\left( {{\tan }^{2}}A \right) \right]-2{{\tan }^{2}}A \)We know that\({{\sec }^{2}}A-{{\tan}^{2}}A=1\)Therefore,\( ={{\sec }^{2}}A+{{\tan}^{2}}A-2{{\tan }^{2}}A\)\( ={{\sec }^{2}}A-{{\tan}^{2}}A\)\( =1 \)Hence, proved