Question

if 13 sin theta minus 12 cos theta = 0 find the value of 2 sin theta cos theta / cos square theta - sin square theta

Answer 1

Answer:

$\frac{2\:sin\theta\:cos\theta}{cos ^2\theta-sin^2\theta}=\frac{312}{25}$

Step-by-step explanation:

$\text{Given:}$

$13\:sin\theta-12\:cos\theta=0$

$\implies\:13\:sin\theta=12\:cos\theta$

$\implies\:\frac{sin\theta}{cos\theta}=\frac{12}{13}$

$\implies\:tan\theta=\frac{12}{13}$

$\text{Now,}$

$\frac{2\:sin\theta\:cos\theta}{cos ^2\theta-sin^2\theta}$

$=\frac{sin\:2\theta}{cos\:2\theta}$

$=tan\:2\theta$

$=tan\:2\theta$

$=\frac{2tan\theta}{1-tan^2\theta}$

$=\frac{2(\frac{12}{13})}{1-(\frac{12}{13})^2}$

$=\frac{\frac{24}{13}}{1-\frac{144}{169}}$

$=\frac{\frac{24}{13}}{\frac{169-144}{169}}$

$=\frac{\frac{24}{13}}{\frac{25}{169}}$

$=\frac{24}{13}*\frac{169}{25}$

$=\frac{24}{1}*\frac{13}{25}$

$=\frac{312}{25}$

$\implies\frac{2\:sin\theta\:cos\theta}{cos ^2\theta-sin^2\theta}=\frac{312}{25}$