A) \[\frac{{{a}^{2}}-{{b}^{2}}}{2ab}\]
B) \[\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]
C) \[\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
D) \[\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}\]
Correct Answer: B
Solution :
\[\sin \theta =\frac{2ab}{{{a}^{2}}+{{b}^{2}}},\]here\[AC={{a}^{2}}+{{b}^{2}}\] BC = 2ab \[\therefore \]\[A{{B}^{2}}=A{{C}^{2}}-B{{C}^{2}}={{({{a}^{2}}+{{b}^{2}})}^{2}}-{{(2ab)}^{2}}\] \[={{a}^{4}}+{{b}^{4}}2{{a}^{2}}{{b}^{2}}-4{{a}^{2}}{{b}^{2}}\] \[={{a}^{4}}+{{b}^{4}}-2{{a}^{2}}{{b}^{2}}={{({{a}^{2}}-{{b}^{2}})}^{2}}\] \[\therefore \]\[AB={{a}^{2}}-{{b}^{2}}\] \[\therefore \]\[\tan \theta =\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]You need to login to perform this action.
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