Question

If cosec (A + B) = 1 and cosec(A – B) = 2, 0° < (A + B) ≤ 90° and A > B then find the values of :
(i) sin A cos B + cos A sin B
(ii) $\frac{\tan A-\tan B}{1+\tan A\tan B}$

Answer 1

As we know that,
$$\begin{gathered} \mathrm{cosec}90°=1 \\ \mathrm{Thus}, \\ \mathrm{if}\mathrm{cosec}\left(A+B\right)=1 \\ \Rightarrow A+B=90°...\left(1\right) \\ \mathrm{and}\mathrm{cosec}30°=2 \\ \mathrm{Thus}, \\ \mathrm{if}\mathrm{cosec}\left(A-B\right)=2 \\ \Rightarrow A-B=30°...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ A=60°\mathrm{and}B=30°...\left(3\right) \\ \mathrm{Now}, \\ \left(\mathrm{i}\right)\sin 60°=\frac{\sqrt{3}}{2} \\ \cos 30°=\frac{\sqrt{3}}{2} \\ \cos 60°=\frac{1}{2} \\ \sin 30°=\frac{1}{2} \\ \mathrm{On}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ \sin A\cos B+\cos A\sin B=\sin 60°\cos 30°+\cos 60°\sin 30° \\ =\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2} \\ =\frac{3}{4}+\frac{1}{4} \\ =\frac{4}{4} \\ =1 \\ \mathrm{Hence},\sin A\cos B+\cos A\sin B=1. \\ \left(\mathrm{ii}\right)\tan 60°=\sqrt{3} \\ \tan 30°=\frac{1}{\sqrt{3}} \\ \mathrm{On}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ \frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\tan 60°-\tan 30°}{1+\tan 60°\tan 30°} \\ =\frac{\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}}}{1+\sqrt{3}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)} \\ =\frac{{\displaystyle \frac{3-1}{\sqrt{3}}}}{1+1} \\ =\frac{2}{2\sqrt{3}} \\ =\frac{1}{\sqrt{3}} \\ \mathrm{Hence},\frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{1}{\sqrt{3}}. \end{gathered}$$