Question

if sinx=5/13,π/2<x<π,findthe value of tanx+secx​

Answer 1

Given →

sin x= 5/13

and x lies in ll quadrant.

To find →

tanx + secx

Solution →

$\cos( {x}^{2} ) = \sqrt{1 - ( { \sin(x ) }^{2} }$

$= > \sqrt{1 - ( { \frac{5}{13} )}^{2} }$

$= > \sqrt{1 - \frac{25}{169} }$

$= > \sqrt{ \frac{169 - 25}{169} }$

$= > \sqrt{ \frac{144}{169} }$

$= > - \frac {12}{13}$

So, cos x = -12/13 as cos functions negative in ll quadrant

as we know sec is the reciprocal of cos , So

sec x = 1/cos x

$= > \sec(x) = - \frac{13}{12}$

Now tanx =>

$= > \tan(x) = \frac{ \sin(x) }{ \cos x}$

$= > \tan(x) = \frac{5}{13} \times - \frac{13}{12}$

$= > \tan(x) = - \frac{5}{12}$

Now according to the question-

tanx +secx =- (5/12 )+(-13/12)

$= > \frac{ - 5 - 13}{12}$

$= > \frac{ - 18}{12}$

$= > - \frac{ 3}{2}$

So , tanx +sec x = -3/2