A) 1
B) 3
C) 4
D) 2
E) 5
Correct Answer: D
Solution :
Given, tan A and tan B are the roots of the equation \[ab{{x}^{2}}-{{c}^{2}}x+ab=0\] \[\therefore \] \[\tan A+\tan B=\frac{{{c}^{2}}}{ab},\tan A\tan B=1\] Now, \[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\] \[=\frac{{{c}^{2}}/ab}{1-1}=\infty \] \[\Rightarrow \] \[A+B=\frac{\pi }{2}\]\[\Rightarrow \] \[C=\frac{\pi }{2}\] \[\therefore \] \[si{{n}^{2}}A+si{{n}^{2}}B+si{{n}^{2}}C\] \[={{\sin }^{2}}\left( \frac{\pi }{2}-B \right)+{{\sin }^{2}}B+{{\sin }^{2}}\frac{\pi }{2}\] \[={{\cos }^{2}}B+{{\sin }^{2}}B+1\] \[=2\]You need to login to perform this action.
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