Question

find the solution :- 2tan theta - cot theta +1 =0​

Answer 1

SOLUTION

$= > 2tan \theta - cot \theta = - 1 \\ = > 2tan \theta - \frac{1}{tan \theta} = - 1 \\ = > let \theta = t \\ = > 2t - \frac{1}{t} = - 1 \\ = > \frac{2 {t}^{2} - 1 }{t} = - 1 \\ = > 2 {t}^{2} - 1 = - t \\ = > 2 {t}^{2} + t - 1 = 0 \\ = > 2 {t}^{2} + 2t - t - 1 = 0 \\ = > 2t(t + 1) - 1(t + 1) = 0 \\ = > ( t + 1)(2t - 1) = 0 \\ = > t = - 1 \: \: \: \: \: o r \: \: \: t = 2t = 1 \\ = > t = - 1 \: \: \: or \: \: \: \: t = \frac{1}{2} \\ \\ = > tan \theta = - 1 \: or \: tan \theta = \frac{1}{2} \\ = > tan \theta = - tan \frac{\pi}{4} \: \: or \: \: tan \theta = tan(tan {}^{ - 1} ( \frac{1}{2} )) \\ = > tan \theta = tan(\pi - \frac{\pi}{4} ) \: \: or \: \: tan \theta = tan( {tan}^{ - 1} ( \frac{1}{2} )) \\ = > tan \theta = tan( \frac{3\pi }{4} ) \: \: or \: \: tan \theta = tan(tan {}^{ - 1} ( \frac{1}{2} ))$

Use general solution for tan theta = tan alpha, theta= nπ + alpha

=) Theta= nπ +3π/4 or theta= mπ+tan^-1

(1/2)

hope it helps ☺️