Question

solve this problem(attach solution if you can)​

Answer 1

Answer:

$\huge{\boxed{\red{\bf{\mathfrak{Your\:solution}}}}}$

sec¢ = ${x}+ \frac{1}{4x}$

by squaring

${sec}^2¢$ = { ${x}+ \frac{1}{4x}$ }^2

we know that ${sec}^2¢ - {tan}^2¢ = 1$

so ${tan}^2¢$ = ${sec}^2¢$ - 1

${tan}^2¢$ = { ${x}+ \frac{1}{4x}$ }^2 - 1

after solving this we get

tan¢ = ${x}+ \frac{1}{4x}$ and ${x} - \frac{1}{4x}$

putting both values

${x}+ \frac{1}{4x}$ + ${x}+ \frac{1}{4x}$

= 1 / 2x

${x}+ \frac{1}{4x}$ + ${x} - \frac{1}{4x}$

= 2x

#answerwithquality and #BAL

Answer 2

Answer:

2x

Explanation:

sec a = hypotensus /adjacent , therefore on taking l c m of sec a we get 4x2+1 /4x therefore by pythagorus therom we can find opposite side p2=h2-b2 . subsitute value of h and b then p =4x2-1 ,tan a =p/b =4x2-1/4x. 2 . take Lhs sec a +tana = 4x2 -1 /4x + 4x2+1/4x = 8x2/4x=2x