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, \[\cot \theta =\frac{b}{a}\] |
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