If $\sin \theta=\frac{-12}{13}, \cos \phi=\frac{-4}{5}$ and $\theta, \phi$ lie in the third quadrant, then…
If $\sin \theta=\frac{-12}{13}, \cos \phi=\frac{-4}{5}$ and $\theta, \phi$ lie in the third quadrant, then $\tan(\theta-\phi)=$
- $\frac{-33}{56}$
- $\frac{-56}{33}$
- $\frac{56}{33}$
- $\frac{33}{56}$
Solution
$\sin \theta=\frac{-12}{13} \Rightarrow \cos \theta=\sqrt{1-\left(\frac{-12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\pm \frac{5}{13}$
$\because \theta$ is in third quadrant $\Rightarrow \cos \theta \quad=-\frac{5}{13} \Rightarrow \frac{\sin \theta}{\cos \theta}=\tan \theta=\frac{12}{5}$
$\cos \phi=-\frac{4}{5} \Rightarrow \sin \phi=\sqrt{1-\left(\frac{-4}{5}\right)^{2}}=\sqrt{\frac{9}{25}}=\pm \frac{3}{5}$
$\phi$ is in third quadrant $\Rightarrow \sin \phi=\frac{-3}{5} \Rightarrow \frac{\sin \phi}{\cos \phi}=\tan \phi=\frac{3}{4}$ $\tan (\theta-\phi)=\frac{\tan \theta-\tan \phi}{1+\tan \phi \cdot \tan \phi}$
$=\frac{\frac{12}{5}-\frac{3}{4}}{1+\left(\frac{12}{5} \times \frac{3}{4}\right)}=\frac{\frac{33}{20}}{\frac{56}{20}}=\frac{33}{56}$
Asked in: MHT CET 2020 (16 Oct Shift 1)
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