KCET · Maths · Three Dimensional Geometry
If \( \sin \theta=\sin \alpha \), then
- A \( \frac{\theta+\alpha}{2} \) is any odd multiple of \( \frac{\pi}{2} \) and \( \frac{\theta-\alpha}{2} \) is any multiple of
- B \( \frac{\theta+\alpha}{2} \) is any even multiple of \( \frac{\pi}{2} \) and \( \frac{\theta-\alpha}{2} \) is any odd multiple of \( I \)
- C \( \frac{\theta+\alpha}{2} \) is any multiple of \( \frac{\pi}{2} \) and \( \frac{\theta-\alpha}{2} \) is any odd multiple of
- D \( \frac{\theta+\alpha}{2} \) is any multiple of \( \frac{\pi}{2} \) and \( \frac{\theta-\alpha}{2} \) is any even multiple of
Answer & Solution
Correct Answer
(A) \( \frac{\theta+\alpha}{2} \) is any odd multiple of \( \frac{\pi}{2} \) and \( \frac{\theta-\alpha}{2} \) is any multiple of
Step-by-step Solution
Detailed explanation
Given \(\sin \theta=\sin \alpha\)
\(\sin \theta-\sin \alpha=0\)
\(2 \cos \left(\frac{\theta+\alpha}{2}\right) \sin \left(\frac{\theta-\alpha}{2}\right)=0\)
It is not necessary that \(\frac{\theta+\alpha}{2}\) is odd multiple of \(\pi / 2\) and \(\frac{\theta-\alpha}{2}\) is any multiple of \(\pi\) should be simultaneous hold good for
the above equation to be true
Hence, the correct answer should be
\(\frac{\theta+\alpha}{2}\) is any odd multiple of \(\pi / 2\)
or \(\frac{\theta-\alpha}{2}\) is any multiple of \(\pi\)
\(\sin \theta-\sin \alpha=0\)
\(2 \cos \left(\frac{\theta+\alpha}{2}\right) \sin \left(\frac{\theta-\alpha}{2}\right)=0\)
It is not necessary that \(\frac{\theta+\alpha}{2}\) is odd multiple of \(\pi / 2\) and \(\frac{\theta-\alpha}{2}\) is any multiple of \(\pi\) should be simultaneous hold good for
the above equation to be true
Hence, the correct answer should be
\(\frac{\theta+\alpha}{2}\) is any odd multiple of \(\pi / 2\)
or \(\frac{\theta-\alpha}{2}\) is any multiple of \(\pi\)
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