Question
If $\sec\theta=\frac{5}{4},$ find the value of $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}.$
✓
Answer
Given: $\sec\theta=\frac{5}{4}\ \dots(1)$
To find the value of $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}$
Now we know that $\sec\theta=\frac{1}{\cos\theta}$
Therefore,
$\cos\theta=\frac{1}{\cos\theta}$
Therefore from equation (1)
$\cos\theta=\frac{1}{\frac{5}{4}}$
$\cos\theta=\frac{4}{5}\ \dots(2)$
Also, We Know that $\cos^2\theta+\sin^2\theta=1$
Therefore,
$\sin^2\theta=1-\cos^2\theta$
$\sin^2\theta=\sqrt{1-\cos^2\theta}$
Substituting the value of $\cos\theta$ from equation (2)
We get,
$\sin^2\theta=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$
$=\sqrt{1-\frac{4^2}{5^2}}$
$=\sqrt{1-\frac{16}{25}}$
$=\sqrt{\frac{25-16}{25}}$
$=\sqrt{\frac{9}{25}}$
$=\frac{3}{5}$
Therefore,
$\sin\theta=\frac{3}{5}\ \dots(3)$
Also, we know that $\sec^2\theta=1+\tan^2\theta.$
Therefore,
$\tan^2\theta=\sec^2-1$
Therefore
$\tan^2\theta=\Big(\frac{5}{4}\Big)^2-1$
$=\frac{25}{16}-1$
$=\frac{9}{16}$
Therefore,
$\tan\theta=\sqrt{\frac{9}{16}}$
$=\frac{3}{4}$
Therefore,
$\tan\theta=\frac{3}{4}\ \dots(4)$
Also $\cot\theta=\frac{1}{\tan\theta}$
Therefore, from equation (4)
We get,
$\cot\theta=\frac{1}{\frac{3}{4}}$
$\cot\theta=\frac{4}{3}\ \dots(5)$
Substitutiing the value of $\cos\theta,\sin\theta,\cot\theta$ and $\tan\theta$ from equation (2) (3) (4) and (5) respectictively in the expression below
$\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}$
We get,
$\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}=\frac{\frac{3}{5}-2\Big(\frac{4}{5}\Big)}{\frac{3}{4}-\frac{4}{3}}$$$
$=\frac{\frac{3}{5}-\frac{8}{5}}{\frac{(3\times3)-(4\times4)}{4\times3}}$
$=\frac{\frac{3-8}{5}}{\frac{9-16}{4\times3}}$
$=\frac{\frac{-5}{5}}{\frac{-7}{12}}$
$=\frac{12}{7}$
Therefore, $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}=\frac{12}{7}$
To find the value of $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}$
Now we know that $\sec\theta=\frac{1}{\cos\theta}$
Therefore,
$\cos\theta=\frac{1}{\cos\theta}$
Therefore from equation (1)
$\cos\theta=\frac{1}{\frac{5}{4}}$
$\cos\theta=\frac{4}{5}\ \dots(2)$
Also, We Know that $\cos^2\theta+\sin^2\theta=1$
Therefore,
$\sin^2\theta=1-\cos^2\theta$
$\sin^2\theta=\sqrt{1-\cos^2\theta}$
Substituting the value of $\cos\theta$ from equation (2)
We get,
$\sin^2\theta=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$
$=\sqrt{1-\frac{4^2}{5^2}}$
$=\sqrt{1-\frac{16}{25}}$
$=\sqrt{\frac{25-16}{25}}$
$=\sqrt{\frac{9}{25}}$
$=\frac{3}{5}$
Therefore,
$\sin\theta=\frac{3}{5}\ \dots(3)$
Also, we know that $\sec^2\theta=1+\tan^2\theta.$
Therefore,
$\tan^2\theta=\sec^2-1$
Therefore
$\tan^2\theta=\Big(\frac{5}{4}\Big)^2-1$
$=\frac{25}{16}-1$
$=\frac{9}{16}$
Therefore,
$\tan\theta=\sqrt{\frac{9}{16}}$
$=\frac{3}{4}$
Therefore,
$\tan\theta=\frac{3}{4}\ \dots(4)$
Also $\cot\theta=\frac{1}{\tan\theta}$
Therefore, from equation (4)
We get,
$\cot\theta=\frac{1}{\frac{3}{4}}$
$\cot\theta=\frac{4}{3}\ \dots(5)$
Substitutiing the value of $\cos\theta,\sin\theta,\cot\theta$ and $\tan\theta$ from equation (2) (3) (4) and (5) respectictively in the expression below
$\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}$
We get,
$\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}=\frac{\frac{3}{5}-2\Big(\frac{4}{5}\Big)}{\frac{3}{4}-\frac{4}{3}}$$$
$=\frac{\frac{3}{5}-\frac{8}{5}}{\frac{(3\times3)-(4\times4)}{4\times3}}$
$=\frac{\frac{3-8}{5}}{\frac{9-16}{4\times3}}$
$=\frac{\frac{-5}{5}}{\frac{-7}{12}}$
$=\frac{12}{7}$
Therefore, $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}=\frac{12}{7}$
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