Question

If $\cot \theta =\frac{a}{b}$ then $\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=$

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{a}{a+b}$
• D. $\frac{a-b}{a+b}$

Answer 1

Answer: (D)

$\frac{a-b}{a+b}$

$\cot \theta =\frac{a}{b}$

By Pythagoras theorem,

hypotenuse $=\sqrt{a^2+b^2}$ then,

$\sin \theta =\frac{b}{\sqrt{a^2+b^2}}$ and $\cos \theta =\frac{a}{\sqrt{a^2+b^2}}$.

$\therefore \frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{\frac{a}{\sqrt{a^2+b^2}}-\frac{b}{\sqrt{a^2+b^2}}}{\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}}=\frac{\frac{a-b}{\sqrt{a^2+b^2}}}{\frac{a+b}{\sqrt{a^2+b^2}}}=\frac{a-b}{a+b}$