question_answer

If sin \[\theta =\frac{2ab}{{{a}^{2}}+{{b}^{2}}},\]then the value of tan 6 is equal to

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 \[\therefore \]\[A{{B}^{2}}=A{{C}^{2}}-B{{C}^{2}}={{({{a}^{2}}+{{b}^{2}})}^{2}}-{{(2ab)}^{2}}\] \[={{a}^{4}}+{{b}^{4}}2{{a}^{2}}{{b}^{2}}-4{{a}^{2}}{{b}^{2}}\] \[={{a}^{4}}+{{b}^{4}}-2{{a}^{2}}{{b}^{2}}={{({{a}^{2}}-{{b}^{2}})}^{2}}\] \[\therefore \]\[AB={{a}^{2}}-{{b}^{2}}\] \[\therefore \]\[\tan \theta =\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]