question_answer

If \[\sin \theta =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}},\]then the value of \[\cot \theta \] will be.

A) \[\frac{b}{a}\]

B) \[\frac{a}{b}\]

C) \[\frac{a}{b}+1\]                      

D) \[\frac{b}{a}+1\]

Correct Answer: A

Solution :

Given, \[\sin \theta =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]   .(i)

We know that, \[\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

Now in \[\Delta \,ABC,\] \[\sin \theta =\frac{AB}{AC}\]      .... (ii)

On comparing Eqs. (i) and (ii), we get

\[AB=a\]and \[AC=\sqrt{{{a}^{2}}+{{b}^{2}}}\]

Now in \[\Delta \,ABC,\] by Pythagoras theorem, we have

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[{{(\sqrt{{{a}^{2}}+{{b}^{2}}})}^{2}}={{(a)}^{2}}+{{(BC)}^{2}}\]

\[{{(BC)}^{2}}={{a}^{2}}+{{b}^{2}}-{{a}^{2}}\]

\[\Rightarrow \]   \[B{{C}^{2}}={{b}^{2}}\Rightarrow BC=b\]

\[\Rightarrow \]   \[\cot \theta =\frac{\text{Base}}{\text{Perpendicular}}=\frac{BC}{AB}\]

On putting values, \[\cot \theta =\frac{b}{a}\]