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# If Sin a = 1 2 , Cos B = 12 13 , Where π 2 < a < π and 3 π 2 < B < 2π, Find Tan (A − B). - CBSE (Science) Class 11 - Mathematics

ConceptTrigonometric Functions of Sum and Difference of Two Angles

#### Question

If $\sin A = \frac{1}{2}, \cos B = \frac{12}{13}$, where $\frac{\pi}{2}$< A < π and $\frac{3\pi}{2}$ < B < 2π, find tan (A − B).

#### Solution

Given:
$\sin A = \frac{1}{2}\text{ and }\cos B = \frac{12}{13}$
$\text{ Here, }\frac{\pi}{2} < A < \pi \text{ and }\frac{3\pi}{2} < B < 2\pi .$
That is, A is in the second quadrant and B is in the fourth quadrant .
We know that in the second quadrant, sine function is positive and cosine and tan functions are negative .
In the fourth quadrant, sine and tan functions are negative and cosine function is positive .
Therefore,
$\cos A = - \sqrt{1 - \sin^2 A} = - \sqrt{1 - \left( \frac{1}{2} \right)^2} = - \sqrt{1 - \frac{1}{4}} = - \sqrt{\frac{3}{4}} = \frac{- \sqrt{3}}{2}$
$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{- \sqrt{3}}{2}} = \frac{- 1}{\sqrt{3}}$
$\sin B = - \sqrt{1 - \cos^2 B} = - \sqrt{1 - \left( \frac{12}{13} \right)^2} = - \sqrt{1 - \frac{144}{169}} = - \sqrt{\frac{25}{169}} = \frac{- 5}{13}$
$\tan B = \frac{\sin B}{\cos B} = \frac{- \frac{5}{13}}{\frac{12}{13}} = \frac{- 5}{12}$
$\text{ Now, }\tan\left( A - B \right) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
$= \frac{\frac{- 1}{\sqrt{3}} - \frac{- 5}{12}}{1 + \frac{- 1}{\sqrt{3}} \times \frac{- 5}{12}}$
$= \frac{\frac{- 12 + 5\sqrt{3}}{12\sqrt{3}}}{\frac{12\sqrt{3} + 5}{12\sqrt{3}}} = \frac{5\sqrt{3} - 12}{5 + 12\sqrt{3}}$

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#### APPEARS IN

RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 7: Values of Trigonometric function at sum or difference of angles
Ex.7.10 | Q: 5 | Page no. 19
Solution If Sin a = 1 2 , Cos B = 12 13 , Where π 2 < a < π and 3 π 2 < B < 2π, Find Tan (A − B). Concept: Trigonometric Functions of Sum and Difference of Two Angles.
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