
Answer
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Hint: The angle \[\alpha \] is in the second quadrant and \[\beta \] is in the third quadrant. Derive the value of \[\cos \alpha \] from the given value of \[\sin \alpha \] and \[\cos \beta \],\[\sin \beta \] from the given value of \[\tan \beta \] .
Then apply the formula of \[\sin (\beta - \alpha )\] and substitute the required values to obtain the required answer.
Formula used:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[\tan \theta = \dfrac{a}{b} \Leftrightarrow \sin \theta = \dfrac{a}{{\sqrt {{a^2} + {b^2}} }},\cos \theta = \dfrac{b}{{\sqrt {{a^2} + {b^2}} }}\]
\[\sin (a - b) = \sin a\cos b - \cos a\sin b\]
Complete step by step solution:
The angle \[\alpha \] is in the second quadrant and \[\beta \] is in the third quadrant.
It is given that \[\sin \alpha = \dfrac{{15}}{{17}}\],
Therefore,
\[\cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } \]
=\[\sqrt {1 - {{\left( {\dfrac{{15}}{{17}}} \right)}^2}} \]
=\[\sqrt {\dfrac{{289 - 225}}{{289}}} \]
=\[\pm \dfrac{8}{{17}}\]
But, as the angle \[\alpha \] is in the second quadrant, so \[\cos \alpha = - \dfrac{8}{{17}}\]
Now, \[\tan \beta = \dfrac{{12}}{5}\]
\[\sin \beta = \dfrac{{12}}{{\sqrt {{5^2} + {{12}^2}} }}\]
\[ = \dfrac{{12}}{{\sqrt {169} }}\]
=\[\dfrac{{12}}{{13}}\]
Hence, \[\cos \beta = \dfrac{5}{{13}}\]
But, as \[\beta \] is in the third quadrant so \[\sin \beta = - \dfrac{{12}}{{13}}\]and \[\cos \beta = - \dfrac{5}{{13}}\].
Now,
\[\sin (\beta - \alpha ) = \sin \beta \cos \alpha - \cos \beta \sin \alpha \]
Substitute \[\sin \beta = - \dfrac{{12}}{{13}}\],\[\cos \beta = - \dfrac{5}{{13}}\],\[\sin \alpha = \dfrac{{15}}{{17}}\] and \[\cos \alpha = - \dfrac{8}{{17}}\],
\[ = \left( { - \dfrac{{12}}{{13}}} \right).\left( { - \dfrac{8}{{17}}} \right) - \left( { - \dfrac{5}{{13}}} \right).\left( {\dfrac{{15}}{{17}}} \right)\]
\[ = \dfrac{{96}}{{221}} + \dfrac{{75}}{{221}}\]
\[ = \dfrac{{171}}{{221}}\]
Therefore, the correct option is D.
Note Sometimes students ignore the range of the angles and just solve the problem; therefore, the solved answer will be incorrect as the signs of sine and cosine change according to the angle. So, we have to pay attention to the given angle for the correct answer.
Then apply the formula of \[\sin (\beta - \alpha )\] and substitute the required values to obtain the required answer.
Formula used:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[\tan \theta = \dfrac{a}{b} \Leftrightarrow \sin \theta = \dfrac{a}{{\sqrt {{a^2} + {b^2}} }},\cos \theta = \dfrac{b}{{\sqrt {{a^2} + {b^2}} }}\]
\[\sin (a - b) = \sin a\cos b - \cos a\sin b\]
Complete step by step solution:
The angle \[\alpha \] is in the second quadrant and \[\beta \] is in the third quadrant.
It is given that \[\sin \alpha = \dfrac{{15}}{{17}}\],
Therefore,
\[\cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } \]
=\[\sqrt {1 - {{\left( {\dfrac{{15}}{{17}}} \right)}^2}} \]
=\[\sqrt {\dfrac{{289 - 225}}{{289}}} \]
=\[\pm \dfrac{8}{{17}}\]
But, as the angle \[\alpha \] is in the second quadrant, so \[\cos \alpha = - \dfrac{8}{{17}}\]
Now, \[\tan \beta = \dfrac{{12}}{5}\]
\[\sin \beta = \dfrac{{12}}{{\sqrt {{5^2} + {{12}^2}} }}\]
\[ = \dfrac{{12}}{{\sqrt {169} }}\]
=\[\dfrac{{12}}{{13}}\]
Hence, \[\cos \beta = \dfrac{5}{{13}}\]
But, as \[\beta \] is in the third quadrant so \[\sin \beta = - \dfrac{{12}}{{13}}\]and \[\cos \beta = - \dfrac{5}{{13}}\].
Now,
\[\sin (\beta - \alpha ) = \sin \beta \cos \alpha - \cos \beta \sin \alpha \]
Substitute \[\sin \beta = - \dfrac{{12}}{{13}}\],\[\cos \beta = - \dfrac{5}{{13}}\],\[\sin \alpha = \dfrac{{15}}{{17}}\] and \[\cos \alpha = - \dfrac{8}{{17}}\],
\[ = \left( { - \dfrac{{12}}{{13}}} \right).\left( { - \dfrac{8}{{17}}} \right) - \left( { - \dfrac{5}{{13}}} \right).\left( {\dfrac{{15}}{{17}}} \right)\]
\[ = \dfrac{{96}}{{221}} + \dfrac{{75}}{{221}}\]
\[ = \dfrac{{171}}{{221}}\]
Therefore, the correct option is D.
Note Sometimes students ignore the range of the angles and just solve the problem; therefore, the solved answer will be incorrect as the signs of sine and cosine change according to the angle. So, we have to pay attention to the given angle for the correct answer.
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