# If θ is an acute angle such that cos θ=35, then sin θ tan θ−12 tan2 θ=

Question:

If $\theta$ is an acute angle such that $\cos \theta=\frac{3}{5}$, then $\frac{\sin \theta \tan \theta-1}{2 \tan ^{2} \theta}=$

(a) $\frac{16}{625}$

(b) $\frac{1}{36}$

(C) $\frac{3}{160}$

(d) $\frac{160}{3}$

Solution:

Given: $\cos \theta=\frac{3}{5}$ and we need to find the value of the following expression

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

We know that: $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$\Rightarrow$ Base $=3$

$\Rightarrow$ Hypotenuse $=5$

$\Rightarrow$ Perpendicular $=\sqrt{(\text { Hypotenuse })^{2}-(\text { Base })^{2}}$

$\Rightarrow$ Perpendicular $=\sqrt{25-9}$

$\Rightarrow$ Perpendicular $=4$

Since $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

and $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

So we find,

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

$=\frac{\frac{4}{5} \times \frac{4}{3}-1}{2 \times\left(\frac{4}{3}\right)^{2}}$

$=\frac{\frac{1}{15}}{\frac{32}{9}}$

$=\frac{3}{160}$

Hence the correct option is $(c)$