A) \[\frac{ab+cd}{\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\]
B) \[\frac{ab}{\sqrt{{{a}^{2}}+{{b}^{2}}}}+\frac{bd}{\sqrt{{{c}^{2}}+{{d}^{2}}}}\]
C) \[\frac{ac+bd}{\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\]
D) \[\frac{ab-cd}{\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\]
E) \[\frac{ab-cd}{\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\]
Correct Answer: C
Solution :
We know that, \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[\therefore \]\[\cos \theta =\frac{\left[ \begin{align} & {{(\sqrt{{{a}^{2}}+{{b}^{2}}})}^{2}}+{{(\sqrt{{{c}^{2}}+{{d}^{2}}})}^{2}} \\ & -{{(\sqrt{{{(a-c)}^{2}}+{{(b-d)}^{2}}})}^{2}} \\ \end{align} \right]}{2\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{\left[ \begin{align} & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}-({{a}^{2}}+{{c}^{2}}-2ac \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{b}^{2}}+{{d}^{2}}-2bd) \\ \end{align} \right]}{2\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{ac+bd}{\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\]You need to login to perform this action.
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