Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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If 5 Sin α = 3 Sin ( α + 2 β ) ≠ 0 , Then Tan ( α + β ) is Equal to - Mathematics

MCQ

If  \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\]  is equal to

 

Options

  • \[2 \tan \beta\]

  • \[3 \tan \beta\]

  • \[4 \tan \beta\]

  • \[6 \tan \beta\]

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Solution

\[4 \tan \beta\]

\[ \text{ We have } , \]

\[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right)\]

\[ \Rightarrow \frac{5}{3} = \frac{\sin \left( \alpha + 2 \beta \right)}{\sin \alpha}\]

\[ \Rightarrow \frac{5 - 3}{5 + 3} = \frac{\sin \left( \alpha + 2 \beta \right) - \sin \alpha}{\sin \left( \alpha + 2 \beta \right) + \sin \alpha} \left( \text{ Using componendo and dividendo } \right)\]

\[ \Rightarrow \frac{2}{8} = \frac{\sin \left( \alpha + 2 \beta \right) - \sin \alpha}{\sin \left( \alpha + 2 \beta \right) + \sin \alpha}\]

\[ \Rightarrow \frac{1}{4} = \frac{2\cos\frac{\alpha + 2 \beta + \alpha}{2}\sin\frac{\alpha + 2 \beta - \alpha}{2}}{2\sin\frac{\alpha + 2 \beta + \alpha}{2}\cos\frac{\alpha + 2 \beta - \alpha}{2}}\]

\[ \Rightarrow \frac{1}{4} = \frac{\cos\left( \alpha + \beta \right) \sin \beta}{\sin\left( \alpha + \beta \right) \cos \beta}\]

\[ \Rightarrow \frac{1}{4} = \cot \left( \alpha + \beta \right) \tan \beta\]

\[ \Rightarrow \frac{1}{4} = \frac{1}{\tan \left( \alpha + \beta \right)}\tan \beta\]

\[ \therefore \tan \left( \alpha + \beta \right) = 4 \tan \beta\]

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Q 15 | Page 44
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