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If Sin a = 12 13 and Sin B = 4 5 , Where π 2 < a < π and 0 < B < π 2 , Find the Following: Cos (A + B) - Mathematics

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प्रश्न

 If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
cos (A + B)

संक्षेप में उत्तर
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उत्तर

 Given:
\[\sin A = \frac{12}{13}\text{ and }\sin B = \frac{4}{5}\]
\[\text{When, }\frac{\pi}{2} < A < \pi\text{ and }0 < B < \frac{\pi}{2}, \]
\[\cos A = - \sqrt{1 - \sin^2 A}\text{ and }\cos B = \sqrt{1 - \sin^2 B}\]
( As cosine function is negative in second qudrant and positive in first quadrant )
\[\Rightarrow \cos A = - \sqrt{1 - \left( \frac{12}{13} \right)^2}\text{ and }\cos B = \sqrt{1 - \left( \frac{4}{5} \right)^2}\]
\[ \Rightarrow \cos A = - \sqrt{1 - \frac{144}{169}}\text{ and }\cos B = \sqrt{1 - \frac{16}{25}}\]
\[ \Rightarrow \cos A = - \sqrt{\frac{25}{169}}\text{ and }\cos B = \sqrt{\frac{9}{25}}\]
\[ \Rightarrow \cos A = \frac{- 5}{13}\text{ and }\cos B = \frac{3}{5}\]
Now,
\[\cos\left( A + B \right) = \cos A \cos B - \sin A \sin B\]
\[ = \frac{- 5}{13} \times \frac{3}{5} - \frac{12}{13} \times \frac{4}{5}\]
\[ = \frac{- 15}{65} - \frac{48}{65}\]
\[ = \frac{- 63}{65}\]

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ १९]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 2.2 | पृष्ठ १९

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