Question
If $\sin\theta=\frac{3}{5},$ evaluate $\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}.$
✓
Answer
Given: $\sin\theta=\frac{3}{5}\ \dots(1)$
$\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}$
To find the value of
Now, we know the following trigonometric identity
$\cos^2+\sin^2\theta=1$
Therefore, by substituting the value of $\sin\theta$ from equation (1),
We get,
$\cos^2\theta+\Big(\frac{3}{5}\Big)^2=1$
Therefore,
$\cos^2\theta=1-\Big(\frac{3}{5}\Big)^2$
$=1-\frac{(3)^2}{(5)^2}$
$=1-\frac{9}{25}$
Now by taking L.C.M
We get,
$\cos^2\theta=\frac{25-9}{25}$
$=\frac{16}{25}$
Therefore by taking square root on both sides
We get,
$\cos\theta=\sqrt{\frac{16}{25}}$
$=\frac{\sqrt{16}}{\sqrt{25}}$
$=\frac{4}{5}$
Therefore,
$\cos\theta=\frac{4}{5}\ \dots(2)$
Now, we know that
$\tan\theta=\frac{\sin\theta}{\cos\theta}$
Therefore by substituting the value of $\sin\theta$ and $\cos\theta$ from equation (1) and (2) respectively
We get,
$\tan\theta=\frac{\frac{3}{5}}{\frac{4}{5}}$
$\tan\theta=\frac{3}{5}\times\frac{5}{4}$
$=\frac{3}{4}$
$\tan\theta=\frac{3}{4}\ \dots(3)$
Also, we know that
$\cot\theta=\frac{1}{\tan\theta}$
Therefore from equation (4),
We get,
$\cot\theta=\frac{1}{\frac{3}{4}}$
$=\frac{4}{3}$
Therefore,
$\cot\theta=\frac{4}{3}\ \dots(4)$
Now, by substituting the value of $\cos\theta,\tan\theta$ and $\cot\theta$ from equation (2), (3) and (4) respectively in the expression below
$\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}$
We get,
$=\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{\frac{4}{5}-\frac{1}{\frac{3}{4}}}{2\times\frac{4}{3}}$
$=\frac{\frac{4}{5}-\frac{4}{3}}{\frac{2\times4}{3}}$
$=\frac{\frac{4\times3}{5\times3}-\frac{4\times5}{3\times5}}{\frac{8}{3}}$
$=\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{\frac{12}{15}-\frac{20}{15}}{\frac{8}{3}}$
$=\frac{\frac{-8}{15}}{\frac{8}{3}}$
$=\frac{-8}{15}\times\frac{3}{8}$
$=\frac{-1}{5}$
Therefore, $\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{-1}{5}$
$\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}$
To find the value of
Now, we know the following trigonometric identity
$\cos^2+\sin^2\theta=1$
Therefore, by substituting the value of $\sin\theta$ from equation (1),
We get,
$\cos^2\theta+\Big(\frac{3}{5}\Big)^2=1$
Therefore,
$\cos^2\theta=1-\Big(\frac{3}{5}\Big)^2$
$=1-\frac{(3)^2}{(5)^2}$
$=1-\frac{9}{25}$
Now by taking L.C.M
We get,
$\cos^2\theta=\frac{25-9}{25}$
$=\frac{16}{25}$
Therefore by taking square root on both sides
We get,
$\cos\theta=\sqrt{\frac{16}{25}}$
$=\frac{\sqrt{16}}{\sqrt{25}}$
$=\frac{4}{5}$
Therefore,
$\cos\theta=\frac{4}{5}\ \dots(2)$
Now, we know that
$\tan\theta=\frac{\sin\theta}{\cos\theta}$
Therefore by substituting the value of $\sin\theta$ and $\cos\theta$ from equation (1) and (2) respectively
We get,
$\tan\theta=\frac{\frac{3}{5}}{\frac{4}{5}}$
$\tan\theta=\frac{3}{5}\times\frac{5}{4}$
$=\frac{3}{4}$
$\tan\theta=\frac{3}{4}\ \dots(3)$
Also, we know that
$\cot\theta=\frac{1}{\tan\theta}$
Therefore from equation (4),
We get,
$\cot\theta=\frac{1}{\frac{3}{4}}$
$=\frac{4}{3}$
Therefore,
$\cot\theta=\frac{4}{3}\ \dots(4)$
Now, by substituting the value of $\cos\theta,\tan\theta$ and $\cot\theta$ from equation (2), (3) and (4) respectively in the expression below
$\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}$
We get,
$=\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{\frac{4}{5}-\frac{1}{\frac{3}{4}}}{2\times\frac{4}{3}}$
$=\frac{\frac{4}{5}-\frac{4}{3}}{\frac{2\times4}{3}}$
$=\frac{\frac{4\times3}{5\times3}-\frac{4\times5}{3\times5}}{\frac{8}{3}}$
$=\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{\frac{12}{15}-\frac{20}{15}}{\frac{8}{3}}$
$=\frac{\frac{-8}{15}}{\frac{8}{3}}$
$=\frac{-8}{15}\times\frac{3}{8}$
$=\frac{-1}{5}$
Therefore, $\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{-1}{5}$
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