Question

$\mathrm{If}\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}\mathrm{and}\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}\frac{\pi }{2}<A,B<\pi ,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\tan A+\tan B\mathrm{is}$

• A. $\frac{\sqrt{3}-\sqrt{5}}{\sqrt{5}}$
• B. $\frac{-\sqrt{3}-\sqrt{5}}{\sqrt{5}}$
• C. $\frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}}$
• D. $\frac{-\sqrt{3}-\sqrt{5}}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{-\sqrt{3}-\sqrt{5}}{\sqrt{5}}$

Here, both the angles A and B lie in 2nd quadrant.
So, only sine of the angle will be positive,
Now, we have $\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}\Rightarrow \sin A=\frac{\sqrt{3}}{2}\sin B\mathrm{and}\frac{cosA}{cosB}=\frac{\sqrt{5}}{2}\Rightarrow \cos A=\frac{\sqrt{5}}{2}\cos B\mathrm{Squaring}\mathrm{and}\mathrm{adding}\mathrm{we}\mathrm{get},{\sin }^{2}A+{\cos }^{2}A=\frac{3}{4}{\sin }^{2}B+\frac{5}{4}{\cos }^{2}B\Rightarrow 1=\frac{3{\sin }^{2}B+5{\cos }^{2}B}{4}\Rightarrow 4=3{\sin }^{2}B+5{\cos }^{2}B\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{identity}{\sin }^{2}B=1-{\cos }^{2}B\mathrm{we}\mathrm{get},2{\cos }^{2}B=1\Rightarrow \cos B=\pm \frac{1}{\sqrt{2}}\mathrm{Since},B\mathrm{lies}\mathrm{in}\mathrm{second}\mathrm{quadrant}\mathrm{thus},\cos B=-\frac{1}{\sqrt{2}}\Rightarrow \mathrm{then},\sin B=\frac{1}{\sqrt{2}}\mathrm{Now}\sin A=\frac{\sqrt{3}}{2\sqrt{2}}\mathrm{Also},\cos A=-\frac{\sqrt{5}}{2\sqrt{2}}\mathrm{Thus},\mathrm{we}\mathrm{get}\tan A=\frac{\sin A}{\cos A}=-\frac{\sqrt{3}}{\sqrt{5}}\mathrm{and},\tan B=-1\mathrm{Thus},\tan A+\tan B=-\frac{\sqrt{3}}{\sqrt{5}}-1\Rightarrow \tan A+\tan B=\frac{-\sqrt{3}-\sqrt{5}}{\sqrt{5}}$