Question

If angle A is divided into two parts such that the tangents of one part is B times the tangent of other ,and C is their difference ,then show that sinA =(B+1/B-1)sinC.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{A}=\mathrm{x}+\mathrm{y} \\ \mathrm{Let}\mathrm{C}=\mathrm{x}-\mathrm{y} \\ \mathrm{Now},\tan \mathrm{x}=\mathrm{B}\tan \mathrm{y} \\ \Rightarrow \frac{\mathrm{B}}{1}=\frac{\tan \mathrm{x}}{\tan \mathrm{y}} \\ \Rightarrow \frac{\mathrm{B}}{1}=\frac{\sin \mathrm{x}.\cos \mathrm{y}}{\cos \mathrm{x}.\sin \mathrm{y}} \\ \Rightarrow \frac{\mathrm{B}+1}{\mathrm{B}-1}=\frac{\sin \mathrm{x}.\cos \mathrm{y}+\cos \mathrm{x}.\sin \mathrm{y}}{\sin \mathrm{x}.\cos \mathrm{y}-\cos \mathrm{x}.\sin \mathrm{y}}\left[\mathrm{Applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo}\right] \\ \Rightarrow \frac{\mathrm{B}+1}{\mathrm{B}-1}=\frac{\sin \left(\mathrm{x}+\mathrm{y}\right)}{\sin \left(\mathrm{x}-\mathrm{y}\right)} \\ \Rightarrow \frac{\mathrm{B}+1}{\mathrm{B}-1}=\frac{\sin \mathrm{A}}{\sin \mathrm{C}} \\ \Rightarrow \sin \mathrm{A}=\left(\frac{\mathrm{B}+1}{\mathrm{B}-1}\right)\sin \mathrm{C}. \end{gathered}$$