Question

if 5cos theta=6sin theta evaluate tan theta ​

Answer 1

Given:-

• $\sf 5 \cos{ \theta} = 6 \sin{ \theta}$

To find:-

• Value of :-

1. $\sf \tan{ \theta}$
2. $\sf \dfrac{ 12 \sin{ \theta} - 3 \cos{ \theta}}{12 \sin{ \theta} + 3 \cos{ \theta}}$

Solution:-

★ $\sf 5 \cos{ \theta} = 6 \sin{ \theta}$

$:\implies\sf \dfrac{ \sin{ \theta}}{ \cos{ \theta}} = \dfrac{5}{6}$

i) $\sf \tan{ \theta}$

We know that,

$\sf \tan{ \theta} = \dfrac{ \sin{ \theta}}{ \cos{ \theta}}$

Then, Value of $\sf \tan{ \theta}$ is,

$\sf \tan{ \theta} = \bf{ \dfrac{5}{6}}$

$\rule{200}3$

ii) $\sf \dfrac{ 12 \sin{ \theta} - 3 \cos{ \theta}}{12 \sin{ \theta} + 3 \cos{ \theta}}$

Divide Numerator and denominator by $\sf \cos{ \theta}$

$:\implies\sf \dfrac{ \dfrac{12 \sin{ \theta} - 3 \cos{ \theta}}{ \cos{ \theta}}}{ \dfrac{12 \sin{ \theta} - 3 \cos{ \theta}}{ \cos{ \theta}}} \\\\ :\implies\sf \dfrac{ 12 \dfrac{ \sin{ \theta}}{ \cos{ \theta}} - 3 \dfrac{ \cos{ \theta}}{ \cos{ \theta}}}{12 \dfrac{ \sin{ \theta}}{ \cos{ \theta}} + 3 \dfrac{ \cos{ \theta}}{ \sin{ \theta}}} \\\\ :\implies\sf \dfrac{ 12 \tan{ \theta} - 3}{12 \tan{ \theta} + 3} \\\\ :\implies\sf \dfrac{ \cancel{12} \times \dfrac{5}{ \cancel{6}} - 3}{ \cancel{12} \times \dfrac{5}{ \cancel{6}} + 3}$

$:\implies\sf \dfrac{2 \times 5 - 3}{10 + 3}$

$:\implies\sf \dfrac{10 - 3}{2 \times 5 + 3}$

$:\implies\sf \bf{ \dfrac{7}{13}}$

$\rule{200}3$