Question
If $\cos\theta=\frac{3}{5},$ show that $\frac{(\sin\theta-\cos\theta)}{2\tan\theta}=\frac{3}{160}.$
✓
Answer
$\cos\theta=\frac{3}{5}\Rightarrow\cos^2\theta=\frac{9}{25}$
$\therefore\sin^2\theta=1-\cos^2\theta=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$
$\Rightarrow\sin\theta=\frac45$
$\Rightarrow\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac43$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\frac34$
Thus,
$\text{L.H.S.}=\frac{(\sin\theta-\cot\theta)}{2\tan\theta}$
$=\frac{\frac{4}{5}-\frac{3}{4}}{2\times\frac{4}{3}}$
$=\frac{\frac{16-15}{20}}{\frac{8}{3}}$
$=\frac{1}{20}\times\frac38$
$=\frac{3}{160}$
$=\text{R.H.S.}$
$\therefore\sin^2\theta=1-\cos^2\theta=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$
$\Rightarrow\sin\theta=\frac45$
$\Rightarrow\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac43$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\frac34$
Thus,
$\text{L.H.S.}=\frac{(\sin\theta-\cot\theta)}{2\tan\theta}$
$=\frac{\frac{4}{5}-\frac{3}{4}}{2\times\frac{4}{3}}$
$=\frac{\frac{16-15}{20}}{\frac{8}{3}}$
$=\frac{1}{20}\times\frac38$
$=\frac{3}{160}$
$=\text{R.H.S.}$
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