Question

If $sin\theta =\frac{3}{5},tan\theta =\frac{1}{2}and\frac{\pi }{2}<\theta <\pi <=\frac{3\pi }{2},$ find the value of 8 $tan\theta -\sqrt{5}sec\varphi$.

Answer 1

We have,

$sin\theta =\frac{3}{5},tan\varphi =\frac{1}{2}and\frac{\pi }{2}<\theta <\pi <=\frac{3\pi }{2}$

$\Rightarrow \theta$ lies in the second quadrant and $\varphi$ lies in the third quadrant.

Now, $si{n}^2\theta +co{s}^2\theta =1$

$\Rightarrow co{s}^2\theta =1-si{n}^2\theta$

$\Rightarrow cos\theta =\pm \sqrt{1-si{n}^2\theta }$

In the $2^{nd}$ qudrant cos \theta is negative and tan \theta is also negative

$\therefore cos\theta =-\sqrt{1-si{n}^2\theta }$

$=-\sqrt{1-(\frac{3}{5}{)}^2}$

$=-\sqrt{1-\frac{9}{25}}$

$=-\sqrt{\frac{16}{25}}$

$=-\frac{4}{5}$

$\Rightarrow cos\theta =-\frac{4}{5}$

and, $tan\theta =\frac{sin\theta }{cos\theta }=\frac{\frac{3}{5}}{\frac{-4}{5}}=-\frac{3}{4}...\left(i\right)$

Now, $se{c}^2\varphi -ta{n}^2\varphi =1$

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

$\Rightarrow sec\varphi =\pm \sqrt{1+ta{n}^2\varphi }$

In the third quadrant $sec\varphi$ is negative.

$sec\varphi =\sqrt{1+(\frac{1}{2}{)}^2}$

$=\sqrt{1+\frac{1}{4}}$

$=-\sqrt{\frac{5}{4}}$

$\Rightarrow sec\varphi =-\frac{\sqrt{5}}{2}$

$\therefore 8tan\varphi -\sqrt{5}sec\varphi$

$=8\times \left(\frac{-3}{4}\right)-\sqrt{5}\times \left(\frac{-\sqrt{5}}{2}\right)$

[by equations (i)and (ii)]

$=-2\times 3+\frac{5}{2}$

$=-6+\frac{5}{2}$

$=\frac{-12+5}{2}=\frac{-7}{2}$

$\therefore 8tan\theta =\sqrt{5}sec\varphi =-\frac{7}{2}$