Question

Find a solution of the equation $se{c}^2\theta$ + $ta{n}^2\theta$ = 3 $tan\theta$

Answer 1

The given equation can be written as

(1 + $ta{n}^2\theta$) + $ta{n}^2\theta$ = 3 $tan\theta$

i.e. 2 $ta{n}^2\theta$ - 3 $tan\theta$ + 1 = 0

i.e. 2 $tan\theta$ ( $tan\theta$ - 1) - 1 ($tan\theta$ - 1) = 0

i.e. (2 $tan\theta$ - 1) ( $tan\theta$ - 1) = 0

$\because$ $tan\theta$ = $\frac{1}{2}$ or = ${26}^{\circ }{34}^{{}^{\prime }}$ (from the table of naturla tangents) or $tan\theta$ = 1

i.e. $\theta$ = ${45}^{\circ }$