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If Tan (A − B) = 1 and Sec (A + B) = 2 √ 3 , the Smallest Positive Value of B is

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Question

If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is

 

Options

  • \[\frac{25 \pi}{24}\]

     

  • \[\frac{19 \pi}{24}\]

     

  • \[\frac{13\pi}{24}\]

     

  • \[\frac{11 \pi}{24}\]

     

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Solution

\[\frac{19 \pi}{24}\]

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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Chapter 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.4 [Page 28]

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R.D. Sharma Mathematics [English] Class 11
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.4 | Q 18 | Page 28

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