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