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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II3 Marks
Question
If $\sin 2 \mathrm{~A}=\lambda \sin 2 \mathrm{~B}$, then prove that $\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}$
✓

Answer

$\begin{aligned}
& \sin 2 A=\lambda \sin 2 \mathrm{~B} \\
& \therefore \quad \frac{\sin 2 A}{\sin 2 \mathrm{~B}}=\frac{\lambda}{1}
\end{aligned}$
By componendo-dividendo, we get
$\begin{array}{ll}
& \frac{\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}}{\sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}}=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \frac{2 \sin \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)}{2 \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \sin \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)}=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \frac{2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})}{2 \cos (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \tan (\mathrm{A}+\mathrm{B}) \cdot \cot (\mathrm{A}-\mathrm{B})=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \frac{\tan (\mathrm{A}+\mathrm{B})}{\tan (\mathrm{A}-\mathrm{B})}=\frac{\lambda+1}{\lambda-1}
\end{array}$

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