Question

If $13\sin \theta =5$ find the value of $\frac{(5sin\theta -2\cos \theta )}{tan\theta }$

Answer 1

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Given : $13\sin \theta =5$

$\sin \theta =\frac{5}{13}$

$\Rightarrow {13}^2=5^2+x^2$

$x^2=169-25$

$x=12$

$\therefore cos\theta =\frac{12}{13}$ , $tan\theta =\frac{5}{12}$

Now, $\frac{5sin\theta -2cos\theta }{tan\theta }=\frac{5\times \frac{5}{13}-2\times \frac{12}{13}}{\frac{5}{12}}$

$\Rightarrow \frac{\frac{25-24}{13}}{\frac{5}{12}}=\frac{1}{13}\times \frac{12}{5}=\frac{12}{65}$