Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}\text{'}\mathrm{s}\mathrm{divide}\mathrm{the}\mathrm{angle}\mathrm{A}\mathrm{in}\mathrm{two}\mathrm{part}{\mathrm{\alpha }}_1\mathrm{and}{\mathrm{\alpha }}_2.\mathrm{Hence}, \\ \mathrm{A}={\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2 \\ \mathrm{and}\mathrm{let}\text{'}\mathrm{s}\mathrm{define}, \\ \mathrm{B}={\mathrm{\alpha }}_2-{\mathrm{\alpha }}_1 \\ \mathrm{and}\tan {\mathrm{\alpha }}_2=\mathrm{x}\tan {\mathrm{\alpha }}_1 \\ \mathrm{which}\mathrm{implies}, \\ \frac{{\mathrm{sin\alpha }}_2}{{\mathrm{cos\alpha }}_2}=\frac{{\mathrm{sin\alpha }}_1}{{\mathrm{cos\alpha }}_1} \\ \mathrm{Now}{\mathrm{sin\alpha }}_2{\mathrm{cos\alpha }}_1={\mathrm{xcos\alpha }}_2{\mathrm{sin\alpha }}_1.....\left(1\right) \\ \mathrm{Now}\mathrm{lets}\mathrm{assume}\mathrm{sinB}=\mathrm{k}\mathrm{sinA} \\ \mathrm{using}\mathrm{the}\mathrm{definition}\mathrm{we}\mathrm{may}\mathrm{write}, \\ \sin \left({\mathrm{\alpha }}_2-{\mathrm{\alpha }}_1\right)=\mathrm{k}\sin \left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2\right) \\ \mathrm{or}{\mathrm{sin\alpha }}_2{\mathrm{cos\alpha }}_1-{\mathrm{cos\alpha }}_2\sin {\mathrm{\alpha }}_1=\mathrm{k}{\mathrm{sin\alpha }}_2{\mathrm{cos\alpha }}_1+{\mathrm{kcos\alpha }}_2{\mathrm{sin\alpha }}_1 \\ \left(1-\mathrm{k}\right){\mathrm{sin\alpha }}_2{\mathrm{cos\alpha }}_1=\left(1+\mathrm{k}\right){\mathrm{cos\alpha }}_2{\mathrm{sin\alpha }}_1 \\ {\mathrm{sin\alpha }}_2{\mathrm{cos\alpha }}_1=\frac{\left(1+\mathrm{k}\right)}{\left(1-\mathrm{k}\right)}{\mathrm{cos\alpha }}_2{\mathrm{sin\alpha }}_1...........\left(2\right) \\ \mathrm{From}\mathrm{equation}1\mathrm{and}2\mathrm{we}\mathrm{get}, \\ \mathrm{x}=\frac{\left(1+\mathrm{k}\right)}{\left(1-\mathrm{k}\right)} \\ \mathrm{x}-\mathrm{kx}=1+\mathrm{k} \\ \mathrm{k}=\frac{\mathrm{x}-1}{\mathrm{x}+1} \\ \mathrm{Hence}\mathrm{we}\mathrm{may}\mathrm{say}\mathrm{that}, \\ \mathrm{sinB}=\frac{\mathrm{x}-1}{\mathrm{x}+1}\mathrm{sinA} \\ \mathrm{or}\mathrm{sinA}=\frac{\mathrm{x}+1}{\mathrm{x}-1}\mathrm{sinB} \end{gathered}$$