Question

It is known that $\sin \beta =\frac{4}{5}$ & $0<\beta <\pi$ then the value of $\frac{\sqrt{3}\sin (\alpha +\beta )-\frac{2}{\cos \frac{\pi }{6}}\cos (\alpha +\beta )}{\sin \alpha }$ is

• A. independent of $\alpha$ for all $\beta$ in $(0,\pi )$
• B. $\frac{5}{\sqrt{3}}$ for $\tan \beta <0$
• C. $\frac{\sqrt{3}(7+24\cot \alpha )}{15}$ for $\tan \beta >0$
• D. none

Answer 1

The correct option is D none

$Lety=\frac{\sqrt{3}\sin (\alpha +\beta )-\frac{2}{\cos \frac{\pi }{6}}\cos (\alpha +\beta )}{\sin \alpha }=\frac{\sqrt{3}\sin (\alpha +\beta )-\frac{4}{\sqrt{3}}\cos (\alpha +\beta )}{\sin \alpha }=\frac{3\sin (\alpha +\beta )-4\cos (\alpha +\beta )}{\sqrt{3}\sin \alpha }CaseI:If\beta liesinIquadranti.e.\tan \beta >0,then,y=\frac{3\sin \alpha \left(\frac{3}{5}\right)+3\cos \alpha \left(\frac{4}{5}\right)-4\cos \alpha \left(\frac{3}{5}\right)+4\left(\frac{4}{5}\right)\sin \alpha }{\sqrt{3}\sin \alpha }=\frac{5\sin \alpha }{\sqrt{3}\sin \alpha }=\frac{5}{\sqrt{3}}CaseII:If\beta liesinIIquadrant,i.e.\tan \beta <0then,y=\left(\frac{\frac{7}{5}\sin \alpha +\frac{24}{5}\cos \alpha }{\sqrt{3}\sin \alpha }\right)=\frac{\sqrt{3}(7+24\cot \alpha )}{15}.$