MHT CET · Maths · Trigonometric Ratios & Identities
If \(\cot \alpha=\frac{1}{2}\) and \(\sec \beta=-\frac{5}{3}\) where a \(\in\left(\pi, \frac{3 \pi}{2}\right)\) and \(\beta \in\left(\frac{\pi}{2}, \pi\right)\), then \(\tan (\alpha+\beta)\) has the value
- A \(\frac{3}{11}\)
- B \(\frac{22}{9}\)
- C \(\frac{9}{11}\)
- D \(\frac{2}{11}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{11}\)
Step-by-step Solution
Detailed explanation
\(\cot \alpha=\frac{1}{2} \Rightarrow \tan \alpha=2\)
\(\sec \beta=\frac{-5}{3} \Rightarrow \tan \beta=-\frac{4}{3}\)
Now, \(\tan (\alpha+\beta)=\frac{\tan \theta+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{2+\left(\frac{-4}{3}\right)}{1-2 \times\left(\frac{-4}{3}\right)}=\frac{\frac{6-4}{3}}{\frac{3+8}{3}}=\frac{2}{11}\)
\(\sec \beta=\frac{-5}{3} \Rightarrow \tan \beta=-\frac{4}{3}\)
Now, \(\tan (\alpha+\beta)=\frac{\tan \theta+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{2+\left(\frac{-4}{3}\right)}{1-2 \times\left(\frac{-4}{3}\right)}=\frac{\frac{6-4}{3}}{\frac{3+8}{3}}=\frac{2}{11}\)
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