$\cot \theta=\frac{3}{4}$ এবং $\cos \theta$ ঋণাত্বক হলে, $\frac{\cot (-\theta)+\operatorname{cosec} \theta}{\cos \theta+\sin (-\theta)}$ এর মান নির্ণয় কর।

Solve:

যেহেতু $\cot \theta=\frac{3}{4} \Rightarrow \tan \theta=\frac{4}{3}$ এবং

$\begin{array}{l} \cos \theta \\ \text { negative; } \\ \therefore \quad \sec \theta=-\sqrt{1+\tan ^{2} \theta}=-\sqrt{1+\frac{16}{9}} \\ =-\sqrt{\frac{25}{9}}=-\frac{5}{3} \\ \therefore \cos \theta=-\frac{3}{5} \text { এবং } \\ \sin \theta=\tan \theta \cos \theta=\frac{4}{3} \times\left(-\frac{3}{5}\right)=-\frac{4}{5} \\ \therefore \operatorname{cosec} \theta=-\frac{5}{4} \\ \text { এখন } \frac{\cot (-\theta)+\operatorname{cosec} \theta}{\cos \theta+\sin (-\theta)}=\frac{-\cot \theta+\operatorname{cosec} \theta}{\cos \theta-\sin \theta} \\ =\frac{-\frac{3}{4}+\left(-\frac{5}{4}\right)}{-\frac{3}{5}-\frac{-4}{5}}=\frac{-3-5}{4} \times \frac{5}{-3+4} \\ =-\frac{40}{4}=-10 \text { (Ans.) } \\ \end{array}$