| Statement I: If A is obtuse angle in \[\Delta ABC\], then tan B tan C > 1. |
| Statement II: In \[\Delta ABC\] \[\tan A=\frac{\tan B+\tan C}{\tan B\tan C-1}\] |
A) Statement I is true. Statement J fin true; Statement B is not a correct explanation for Statement I.
B) Statement I is true. Statement II is false.
C) Statement 1 is false. Statement S is true.
D) Statement I is true, Statement H is true; Statement H is a correct explanation for Statement I.
Correct Answer: C
Solution :
Since, \[A+B+C=180{}^\circ \] \[\Rightarrow \] \[A={{180}^{o}}-(B+C)\] \[\therefore \]\[\tan A=\tan [{{180}^{o}}-(B+C)]\] \[=-\tan (B+C)\] \[=-\left\{ \frac{\tan B+\tan C}{1-\tan B\tan C} \right\}\] \[=\frac{\tan B+\tan C}{\tan B\tan C-1}\]Since, A is obtuse. \[\therefore \]\[\tan A<0,\]then\[\tan B+\tan C>0\] \[\therefore \]\[\tan B\tan C-1<0\]\[\Rightarrow \]\[\tan B\tan C<1\] Hence, Statement I is false and Statement II is true.
You need to login to perform this action.
You will be redirected in
3 sec
