question_answer

If \[\sin \theta =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}},\] then the value of cot 6 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, \[\Delta ABC,\]\[\sin \theta =\frac{AB}{AC}\]                                   .(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,

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

\[B{{C}^{2}}={{b}^{2}}\]\[\Rightarrow \]\[BC=b\]\[\Rightarrow \]\[\cot \theta =\frac{b}{a}\]