MHT CET · Maths · Trigonometric Equations
If \(\cot (A+B)=0\), then \(\sin (A+2 B)\) is equal to
- A \(\sin 2 A\)
- B \(\cos A\)
- C \(\sin A\)
- D \(\cos 2 A\)
Answer & Solution
Correct Answer
(C) \(\sin A\)
Step-by-step Solution
Detailed explanation
\(\cot (A+B)=0 \Rightarrow A+B=(2 n+1) \frac{\pi}{2} \Rightarrow B=(2 n\)\(+1) \frac{\pi}{2}-A\)
Now, \(\sin (A+2 B)=\sin (A+(2 n+1) \pi-2 A)\)
\(=\sin ((2 n+1) \pi-A)=\sin A\)
Now, \(\sin (A+2 B)=\sin (A+(2 n+1) \pi-2 A)\)
\(=\sin ((2 n+1) \pi-A)=\sin A\)
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