Question

If $\frac{\sin \alpha }{\sin \beta }=\frac{\sqrt{3}}{2}$ and $\frac{\cos \alpha }{\cos \beta }=\frac{\sqrt{5}}{2},0<\alpha ,\beta <\frac{\pi }{2}$, then

• A. $\tan \alpha =\frac{\sqrt{3}}{\sqrt{5}},\tan \beta =1$
• B. $\tan \alpha =1,\tan \beta =\frac{\sqrt{3}}{\sqrt{5}}$
• C. $\tan \alpha =\frac{\sqrt{5}}{\sqrt{3}},\tan \beta =1$
• D. none of these

Answer 1

The correct option is A $\tan \alpha =\frac{\sqrt{3}}{\sqrt{5}},\tan \beta =1$

$\frac{\sin \alpha }{\sin \beta }=\frac{\sqrt{3}}{2}$

Or $\sqrt{\frac{1-{\cos }^2\alpha }{1-{\cos }^2\beta }}=\frac{\sqrt{3}}{2}$

$\frac{1-{\cos }^2\alpha }{1-{\cos }^2\beta }=\frac{3}{4}$

Or $4-4{\cos }^2\alpha =3-3{\cos }^2\beta$

$4{\cos }^2\alpha -1=3{\cos }^2\beta$ ...(i)

And it is given that

$\cos \alpha =\frac{\sqrt{5}}{2}\cos \beta$

Or ${\cos }^2\alpha =\frac{5}{4}{\cos }^2\beta$

Substituting in equation i, we get

$5co{s}^2\beta -1=3co{s}^2\beta$

Or $2co{s}^2\beta =1$

Or $cos\beta =\frac{1}{\sqrt{2}}=sin\beta$

Therefore

$tan\beta =1$

Now

$cos\beta =sin\beta =\frac{1}{\sqrt{2}}$

Hence

$cos\alpha =\frac{\sqrt{5}}{2\sqrt{2}}$

And $sin\alpha =\frac{\sqrt{3}}{2\sqrt{2}}$

Therefore

$tan\alpha =\frac{sin\alpha }{cos\alpha }=\frac{\sqrt{3}}{\sqrt{5}}$