Question

If $secA=x+\frac{1}{4x}$ then prove that $secA+tanA=2xor\frac{1}{2x}$

Answer 1

Let,

$secA+tanA=y$ ..............1

Then,

$=se{c}^{2}A-ta{n}^{2}A=1$

$=(secA-tanA)(secA+tanA)=1$

$=y(secA-tanA)=1$

$=secA-tanA=\frac{1}{y}$ ....................2

Adding eq. 1 and eq. 2,we get

$=2secA=y+\frac{1}{y}$

$=2(x+\frac{1}{4x})=y+\frac{1}{y}$

$=2x+\frac{1}{2x}=y+\frac{1}{y}$

$=y=2x$ or $y=\frac{1}{2x}$ (on comparing two sides )

$secA+tanA=2x,\frac{1}{2x}$