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प्रश्न
If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is
पर्याय
- \[\frac{25 \pi}{24}\]
- \[\frac{19 \pi}{24}\]
- \[\frac{13\pi}{24}\]
- \[\frac{11 \pi}{24}\]
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उत्तर
Given:
\[\tan(A - B) = 1\text{ and }\sec(A + B) = \frac{2}{\sqrt{3}}\]
\[ \Rightarrow A - B = \frac{\pi}{4} . . . (1)\text{ and }A + B = \frac{\pi}{6} . . . (2)\]
Adding these equations we get:
\[ 2A = \frac{\pi}{4} + \frac{\pi}{6}\]
\[ \Rightarrow A = \frac{5\pi}{24}\]
\[ \Rightarrow\text{ Smallest possible value of B }= \pi - \frac{5\pi}{24} = \frac{19\pi}{24} . \]
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