Question

If sec $\theta =\frac{5}{4}$, find the value of $\frac{sin\theta -2cos\theta }{tan\theta -cot\theta }$

Answer 1

Given: $sec\theta =\frac{5}{4}\because weknowthatcos\theta =\frac{1}{sec\theta }\therefore cos\theta =\frac{4}{5}$ ---(i)

Now, we know that
$co{s}^2\theta +si{n}^2\theta =1$
we can re-write it as,
$sin\theta =\sqrt{1-co{s}^2\theta }$
Substituting the value of $cos\theta$ from equation (i)
we get,
$sin\theta =\sqrt{1-(\frac{4}{5}{)}^2}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{25-16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

$\therefore sin\theta =\frac{3}{5}$ ---(ii)

We also know that,
$se{c}^2\theta =1+ta{n}^2\theta \therefore ta{n}^2\theta =se{c}^2\theta -1\because weknowsec\theta =\frac{5}{4}$
putting the value of $sec\theta$ in the above equation, we get
$ta{n}^2\theta =(\frac{5}{4}{)}^2-1\Rightarrow ta{n}^2\theta =\frac{25}{16}-1\Rightarrow tan\theta =\sqrt{\frac{25-16}{16}}\Rightarrow tan\theta =\sqrt{\frac{9}{16}}\therefore tan\theta =\frac{3}{4}$ ---(iii)

$\because cot\theta =\frac{1}{tan\theta }\therefore cot\theta =\frac{4}{3}$ ---(iv)

substituting the values of $sin\theta$, $cos\theta$, $tan\theta$ and $cot\theta$ from equations (i), (ii), (iii) and (iv) in the equation
$\frac{sin\theta -2cos\theta }{tan\theta -cot\theta }$
we get,
$=\frac{\frac{3}{5}-2\times \left(\frac{4}{5}\right)}{\frac{3}{4}-\frac{4}{3}}=\frac{\frac{3}{5}-\frac{8}{5}}{\frac{3}{4}-\frac{4}{3}}$
Upon simplification we get,
$=\frac{\frac{-5}{5}}{\frac{9-16}{12}}=\frac{12}{7}$