Question

plz help me in solving this

Answer 1

${sec}^{2} \alpha = {x}^{2} +( 1 \div 2 ) + (1 \div 16 {x}^{2} \\ {sec }^{2} \alpha = 1 + {tan}^{2} \alpha \\ {tan}^{2} \alpha = {sec}^{2} \alpha - 1 \\ {tan}^{2} \alpha = {x}^{2} + (1 \div 16 {x}^{2} ) - (1 \div 2) \\ \tan( \alpha ) = x - (1 \div 4x)$
sec x +tan x =2x