MHT CET · Maths · Trigonometric Equations
Let \(\cos (\alpha+\beta)=\frac{4}{5}\) and \(\sin (\alpha-\beta)=\frac{5}{13}\), where \(0 \leq \alpha, \beta \leq \frac{\pi}{4}\), then \(\tan 2 \alpha=\)
- A \(\frac{19}{12}\)
- B \(\frac{56}{33}\)
- C \(\frac{25}{16}\)
- D \(\frac{20}{7}\)
Answer & Solution
Correct Answer
(B) \(\frac{56}{33}\)
Step-by-step Solution
Detailed explanation
\(\cos (\alpha+\beta)=\frac{4}{5} \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \)
\( \sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \)
\( \tan 2 \alpha=\tan \{(\alpha+\beta)+(\alpha-\beta)\}=\)\(\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)} \)
\(=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}=\frac{56}{33}\)
\( \sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \)
\( \tan 2 \alpha=\tan \{(\alpha+\beta)+(\alpha-\beta)\}=\)\(\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)} \)
\(=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}=\frac{56}{33}\)
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