MHT CET · Maths · Trigonometric Ratios & Identities
If \(\sin A=n \sin (A+2 B)\), then \(\tan (A+B)=\)
- A \(\frac{1+n}{2-n} \cdot \tan B\)
- B \(\frac{1-n}{1+n} \cdot \tan B\)
- C \(\frac{1-n}{2+n} \cdot \tan B\)
- D \(\frac{1+n}{1-n} \cdot \tan B\)
Answer & Solution
Correct Answer
(D) \(\frac{1+n}{1-n} \cdot \tan B\)
Step-by-step Solution
Detailed explanation
\(\frac{\sin A}{\sin (A+2 B)} = n\) \(\frac{\sin A + \sin (A+2 B)}{\sin A - \sin (A+2 B)} = \frac{n+1}{n-1}\)
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