Question

If $4\cot \mathrm{\theta }=3\mathrm{then}\left(\frac{\sin \mathrm{\theta }-\cos \mathrm{\theta }}{\sin \mathrm{\theta }+\cos \mathrm{\theta }}\right)=?$
(a) $\frac{3}{7}$

(b) $\frac{2}{7}$

(c) $\frac{1}{7}$

(d) 0

Answer 1

$$\begin{gathered} \mathrm{Given}:4\cot \theta =3 \\ \Rightarrow \cot \theta =\frac{3}{4} \\ \mathrm{Since},\cot \theta =\frac{B}{P} \\ \Rightarrow P=4\mathrm{and}B=3 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 4^2+3^2=H^2 \\ \Rightarrow H^2=16+9 \\ \Rightarrow H^2=25 \\ \Rightarrow H=5 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{4}{5} \\ \cos \theta =\frac{B}{H}=\frac{3}{5} \\ \left(\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }\right)=\left(\frac{{\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}+{\displaystyle \frac{3}{5}}}\right) \\ =\left(\frac{{\displaystyle \frac{4-3}{5}}}{{\displaystyle \frac{4+3}{5}}}\right) \\ =\frac{1}{7} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$