• # question_answer If $4\text{ }tan\text{ }\theta =3$, evaluate $\left( \frac{4\,\,\sin \,\,\theta -\cos \,\,\theta +1}{4\,\,\sin \,\,\theta +\cos \,\,\theta -1} \right)$ OR If tan $2A=cot\text{ (}A-18{}^\circ )$, where 2A is an acute angle, find the value of A.

 Given, $4\text{ }tan\text{ }\theta =3$, $\Rightarrow$ $\tan \,\theta =\frac{3}{4}\left( =\frac{P}{B} \right)$ $P=3K,B=4K,$ Now,              $H=\sqrt{{{P}^{2}}+{{B}^{2}}}$ $=\sqrt{{{(3K)}^{2}}+{{(4K)}^{2}}}$ $=\sqrt{9{{K}^{2}}+16{{K}^{2}}}$ $=\sqrt{25{{K}^{2}}}$ $\Rightarrow$            $H=5K$ $\therefore$      $\sin \,\theta =\frac{P}{H}=\frac{3K}{5K}=\frac{3}{5}$ And      $\cos \,\theta =\frac{B}{H}=\frac{4K}{5K}=\frac{4}{5}$ Now, $\frac{4\,\sin \,\theta -\cos \theta +1}{4\,\sin \,\theta +\cos \,\theta -1}=\frac{4\times \frac{3}{5}-\frac{4}{5}+1}{4\times \frac{3}{5}+\frac{4}{5}-1}$ $=\frac{\left( \frac{12}{5}-\frac{4}{5}+1 \right)}{\left( \frac{12}{5}+\frac{4}{5}-1 \right)}$ $=\frac{\left( \frac{12-4+5}{5} \right)}{\left( \frac{12+4-5}{5} \right)}$ $=\frac{13/5}{11/5}$ $=\frac{13}{11}$ OR Given    $\tan \,2A=cot\,(A-18{}^\circ )$ $\Rightarrow$   $\cot \,(90{}^\circ -2A)=cot(A-18{}^\circ )$ [$\because$ $\tan \,\theta =\cot (90{}^\circ -\theta )$] $\Rightarrow$   $90{}^\circ -2A=A-18{}^\circ$ $\Rightarrow$   $90{}^\circ +18{}^\circ =A+2A$ $\Rightarrow$   $180{}^\circ =3A$ $\Rightarrow$   $A=\frac{108{}^\circ }{3}$ $\Rightarrow$   $A=36{}^\circ$