Math, asked by singareddyraghavared, 3 months ago

Sinx+Cosx= I÷5.
, find tanx​

Answers

Answered by LaeeqAhmed
0

 \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!!

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