Question

Sinx+Cosx= I÷5.
, find tanx​

Answer 1

$\sin(x) + \cos(x) = \frac{1}{5}$

$\bf squaring \: on \: both \: sides :$

${( \sin x + \cos x )}^{2} = ({\frac{1}{5} }) ^{2}$

$\sin ^{2} (x) + \cos^{2} (x) + 2 \sin(x) \cos(x) = \frac{1}{25}$

$1 + \sin(2x) = \frac{1}{25}$

$\sin(2x) = \frac{1}{25} - 1$

$\sin(2x) = \frac{ - 24}{5}$

$\bf similallarly$

$\tan(2x) = \frac{ - 24}{7}$

$\bf but$

$\blue{ \boxed{ \tan(2x) = \frac{2 \tan(x) }{1 - \tan ^{2} (x) } }}$

$\implies \frac{ - 24}{7} = \frac{2 \tan(x) }{1 - \tan^{2} (x) }$

$\implies - 24 + 24 \tan ^{2} (x) = 17 \tan(x)$

$\implies 12 \tan ^{2} (x) + 7 \tan(x) - 12 = 0$

$\bf we \: know \: that :$

$\blue{ \boxed{\bf roots = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} }}$

$\tan(x) = \frac{ - 12± \sqrt{ {(7)}^{2} - 4(12)( - 12) } }{2(12)}$

$\tan(x) = \frac{ - 12± \sqrt{ 49 + 576 } }{2(12)}$

$\tan(x) = \frac{ - 12± \sqrt{ 625} }{24}$

$\tan(x) = \frac{ - 7±25 }{24}$

$\boxed{ \bf \orange{ \tan(x) = \frac{ 3}{4} \: \: \: (or) \: \: \frac{ - 4}{3} }}$

HOPE THIS HELPS!!