Question

If $sec\theta =\frac{5}{4}$, show that $\frac{(sin\theta -2cos\theta )}{(tan\theta -cot\theta )}=\frac{12}{7}$.

Answer 1

Given: $sec\theta =\frac{5}{4}----\left(1\right)$

To find the value of $\frac{sin\theta –2cos\theta }{tan\theta –cot\theta }$

Now we know that $sec\theta =\frac{1}{cos\theta }$

Therefore,

$cos\theta =\frac{1}{sec\theta }$

Therefore from equation (1)

$cos\theta =\frac{4}{5}cos\theta =\frac{4}{5}----\left(2\right)$

Also, we know that $co{s}^2\theta +si{n}^2\theta =1$

Therefore,

$si{n}^2\theta =1–co{s}^2\theta sin\theta =\sqrt{1–co{s}^2\theta }$

Substituting the value of $cos\theta$ from equation (2)

We get,

$sin\theta =\sqrt{1-(\frac{4}{5}{)}^2}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

Therefore,

$sin\theta =\frac{3}{5}----\left(3\right)$

Also, we know that

$se{c}^2\theta =1+ta{n}^2\theta$

Therefore,

$ta{n}^2\theta =(\frac{5}{4}{)}^2-1tan\theta =\sqrt{\frac{9}{16}}$

Therefore,

$tan\theta =\frac{3}{4}----\left(4\right)$

Also, $cot\theta =\frac{1}{tan\theta }$

Therefore from equation (4)

We get,

$cot\theta =\frac{4}{3}----\left(5\right)$

Substituting the value of $cos\theta ,cot\theta$ and $tan\theta$ from the equation (2), (3), (4) and (5) respectively in the expression below.

$\frac{sin\theta –2cos\theta }{tan\theta –cot\theta }$

We get,

$\frac{sin\theta –2cos\theta }{tan\theta –cot\theta }=\frac{\frac{3}{5}-2\left(\frac{4}{5}\right)}{\frac{3}{4}–\frac{4}{3}}=\frac{12}{7}$

Therefore, $\frac{sin\theta –2cos\theta }{tan\theta –cot\theta }=\frac{12}{7}$