Question

If the tangents of the angles A and B of a triangle ABC satisfy the equation $ab{x}^2-c^{2}x+ab=0$, then

• A. $\tan A=\frac{a}{b}$
• B. $\tan B=\frac{b}{a}$
• C. $\cos C=0$
• D. ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\tan A=\frac{a}{b}$
B ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$
C $\tan B=\frac{b}{a}$
D $\cos C=0$

$\tan A+\tan B=\frac{c^2}{ab}$
And
$\tan A.\tan B=1$
Or
$1-\tan A.\tan B=0$
Implies
$\tan (A+B)=\infty$
Or
$C=\frac{\pi }{2}$
Hence
$\tan A=cotB$
Hence $\tan A=\frac{a}{b}$ then $\tan B=\frac{b}{a}$.
Now
${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$
$=1+{\sin }^{2}A+{\sin }^{2}B$
$=1+{\sin }^{2}A+{\cos }^{2}A$ ... $(C={90}^0)$.
$=2$