Question

If 3 tan A = 4 then prove that
(i) $\sqrt{\frac{\sec A-\mathrm{cosec}A}{\sec A+\mathrm{cosec}A}}=\frac{1}{\sqrt{7}}$
(ii) $\sqrt{\frac{1-\sin A}{1+\cos A}}=\frac{1}{2\sqrt{2}}$

Answer 1

(i)
​$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{\sec \theta -\mathrm{cosec}\theta }{\sec \theta +\mathrm{cosec}\theta }} \\ =\sqrt{\frac{\left({\displaystyle \frac{1}{\cos \theta }}-{\displaystyle \frac{1}{\sin \theta }}\right)}{\left({\displaystyle \frac{1}{\cos \theta }}+{\displaystyle \frac{1}{\sin \theta }}\right)}} \\ =\sqrt{\frac{\left({\displaystyle \frac{\sin \theta -\cos \theta }{\sin \theta \cos \theta }}\right)}{\left({\displaystyle \frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }}\right)}} \\ =\sqrt{\frac{\left({\displaystyle \frac{\sin \theta -\cos \theta }{\sin \theta }}\right)}{\left({\displaystyle \frac{\sin \theta +\cos \theta }{\sin \theta }}\right)}} \\ =\sqrt{\frac{\left({\displaystyle \frac{\sin \theta }{\sin \theta }-\frac{{\displaystyle \cos \theta }}{\sin \theta }}\right)}{\left({\displaystyle \frac{\sin \theta }{\sin \theta }}+{\displaystyle \frac{\cos \theta }{\sin \theta }}\right)}} \\ =\sqrt{\frac{1-\cot \theta }{1+\cot \theta }} \\ =\sqrt{\frac{\left(1-{\displaystyle \frac{3}{4}}\right)}{\left(1+{\displaystyle \frac{3}{4}}\right)}} \end{gathered}$$
$$\begin{gathered} =\sqrt{\frac{\left({\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{7}{4}}\right)}} \\ =\sqrt{\frac{1}{7}} \\ =\frac{1}{\sqrt{7}} \\ =\mathrm{RHS} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Given}:3\tan A=4 \\ \Rightarrow \tan A=\frac{4}{3} \\ \mathrm{Since},\tan A=\frac{P}{B} \\ \Rightarrow P=4\mathrm{and}B=3 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 4^2+3^2=H^2 \\ \Rightarrow H^2=16+9 \\ \Rightarrow H^2=25 \\ \Rightarrow H=5 \\ \mathrm{Therefore}, \\ \sin A=\frac{P}{H}=\frac{4}{5} \\ \cos A=\frac{B}{H}=\frac{3}{5} \\ \mathrm{Now}, \\ \sqrt{\frac{1-\sin A}{1+\cos A}}=\sqrt{\frac{{\displaystyle 1-\frac{4}{5}}}{1+{\displaystyle \frac{3}{5}}}} \\ =\sqrt{\frac{{\displaystyle \frac{5-4}{5}}}{{\displaystyle \frac{5+3}{5}}}} \\ =\sqrt{\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{8}{5}}}} \\ =\sqrt{\frac{{\displaystyle 1}}{{\displaystyle 8}}} \\ =\frac{1}{2\sqrt{2}} \\ \mathrm{Hence},\sqrt{\frac{1-\sin A}{1+\cos A}}=\frac{1}{2\sqrt{2}}. \end{gathered}$$