Question

If the tangents of the angles A and B of 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 $\tan B=\frac{b}{a}$
C $\cos C=0$
D ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$

From the given equation, we get
$\tan A+\tan B=\frac{c^2}{ab}$ and $\tan A\tan B=1$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\Rightarrow A+B=\frac{\pi }{2}$ and $C=\frac{\pi }{2}$
Therefore triangle ABC is right angled at C
Hence $\tan A=\frac{a}{b},\tan B=\frac{b}{a},\cos C=0,$
$\sin A=\frac{a}{c},\sin B=\frac{b}{c},\sin C=1$
${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{a^2}{c^2}+\frac{b^2}{c^2}+1$
$=\frac{a^2+b^2}{c^2}+1=1+1=2$