If theta is an acute angle such that costheta=frac{3}{5}, then frac{sinthetatantheta-1}{2tan^2theta}=

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

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: $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$ We know that: $\Rightarrow \text{Base}=3$ $\Rightarrow\text{Hypotenuse}=5$ $\Rightarrow\text{Perpendicular}=\sqrt{(\text{Hypotenuse)}^2-(\text{Base)}^2}$ $\Rightarrow\text{Perpendicular}=\sqrt{25-9}$ $\Rightarrow\text{Perpendicular}=4$ Since $\sin \theta=\frac{\text{perpendicular}}{\text{Hypotenuse}}$ and $\tan \theta=\frac{\text{perpendicular}}{\text{Base}}$ So we find, $\frac{\sin\theta\tan\theta-1}{2\tan^2\theta}$ $=\frac{\frac{4}{5}\times\frac{4}{3}-1}{2\times\Big(\frac{4}{3}\Big)^2}$ $=\frac{\frac{16}{15}-1}{\frac{32}{9}}$ $=\frac{\frac{1}{15}}{\frac{32}{9}}$ $=\frac{3}{160}$ Hence the correct option is $(c)$

M.C.Q (1 Marks)

GSEB - English Medium

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$\frac{160}{3}$

STD 10
Maths
Trigonometric Ratios
R.D. Sharma

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