Question

if $\frac{1}{\cos \theta }=x+\frac{1}{4x},$ then the value of $(\tan \theta +\frac{1}{\cos \theta })$ is

• A. x
• B. 2x
• C. 3x
• D. 4x

Answer 1

The correct option is B 2x

$secA=x+\frac{1}{4x}$
$\frac{H}{B}=\frac{4x^2+1}{4x}$
Thus, $H=4x^2+1,B=4x$
Now, $H^2=P^2+B^2$ (Pythagoras Theorem)
$(4x^2+1{)}^2=P^2+(4x{)}^2$
$16x^4+1+8x^2=P^2+16x^2$
$16x^4-8x^2+1=P^2$
$P=4x^2-1$ or $P=1-4x^2$
When $P=4x^2-1$
Now, $secA+\tan A=\frac{H}{B}+\frac{P}{B}$
= $\frac{4x^2+1}{4x}+\frac{4x^2-1}{4x}$
= $\frac{8x^2}{4x}$
= $2x$
When , $P=1-4x^2$
Now, $secA+\tan A=\frac{H}{B}+\frac{P}{B}$
= $\frac{4x^2+1}{4x}+\frac{-4x^2+1}{4x}$
= $\frac{2}{4x}$
= $\frac{1}{2x}$