Ikunupe6v

2022-01-15

Calculate $\frac{3\mathrm{sin}\alpha +2\mathrm{tan}\alpha }{\mathrm{cot}\alpha }$ if $\mathrm{cos}\alpha =\frac{1}{3}$

RizerMix

$\frac{3\mathrm{sin}\alpha +2\mathrm{tan}\alpha }{\mathrm{cot}\alpha }=\frac{3{\mathrm{sin}}^{2}\alpha \mathrm{cos}\alpha +2{\mathrm{sin}}^{2}\alpha }{{\mathrm{cos}}^{2}\alpha }=\frac{3\left(1-{\mathrm{cos}}^{2}\alpha \right)\mathrm{cos}\alpha +2\left(1-{\mathrm{cos}}^{2}\alpha \right)}{{\mathrm{cos}}^{2}\alpha }$ You should be able to conclude.

Vasquez

$\frac{3\mathrm{sin}x+2\mathrm{tan}x}{\mathrm{cot}x}=\left(3\mathrm{sin}x+2\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)\cdot \frac{\mathrm{sin}x}{\mathrm{cos}x}=\frac{3{\mathrm{sin}}^{2}x}{\mathrm{cos}x}+\frac{2{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=$ $=\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}\left(3\mathrm{cos}x+2\right)=\frac{1-{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}\left(3\mathrm{cos}x+2\right)=\frac{1-\frac{1}{9}}{\frac{1}{9}}\left(3\cdot \frac{1}{3}+2\right)=24$

user_27qwe

Hint : $\frac{3\mathrm{sin}\alpha +2\mathrm{tan}\alpha }{\mathrm{cot}\alpha }=\frac{3\mathrm{sin}\alpha }{\mathrm{cot}\alpha }+\frac{2\mathrm{tan}\alpha }{\mathrm{cot}\alpha }$ But