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. |
Answer:
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 \[\tan \,2A=cot\,(A-18{}^\circ )\] \[\Rightarrow \] \[\cot \,(90{}^\circ -2A)=cot(A-18{}^\circ )\] [\[\because \] \[\tan \,\theta =\cot (90{}^\circ -\theta )\]] \[\Rightarrow \] \[90{}^\circ -2A=A-18{}^\circ \] \[\Rightarrow \] \[90{}^\circ +18{}^\circ =A+2A\] \[\Rightarrow \] \[180{}^\circ =3A\] \[\Rightarrow \] \[A=\frac{108{}^\circ }{3}\] \[\Rightarrow \] \[A=36{}^\circ \]
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