Question

If $1+4x^2=4x\sec A$ then show that $\sec A+\tan A=\frac{1}{2x}$

Answer 1

Given that:

$1+4x^2=4x\sec A$,

$\sec A=\frac{1+4x^2}{4x}$ ......(1)

Now we can calculate perpendicular side of the triangle,

Hypotenuse$=(1+4x^2)$, Base$=4x$

$Perpendicula{r}^2=Hypotenus{e}^2-Bas{e}^2$

$Perpendicula{r}^2=(1+4x^2{)}^2-(4x{)}^2$

$Perpendicula{r}^2=1+16x^4+8x^2-16x^2$

$Perpendicula{r}^2=1+16x^4-8x^2$

$Perpendicula{r}^2=(1-4x^2{)}^2$

$Perpendicular=(1-4x^2)$

So,

$\tan A=\frac{Perpendicular}{Base}$

$\tan a=\frac{1-4x^2}{4x}$ ......(2)

$LHS=\sec A+\tan A$

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

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

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

Hence, $\sec A+\tan A=\frac{1}{2x}$