2022-03-20

Find the solution of cos B csc B tan = 1

Vasquez

$\mathrm{cos}\left(B\right)\mathrm{csc}\left(B\right)\mathrm{tan}\left(x\right)=1$

Write the problem as a mathematical expression.

$\mathrm{cos}\left(B\right)\mathrm{csc}\left(B\right)\mathrm{tan}\left(x\right)=1$

Simplify $\mathrm{cos}\left(B\right)\mathrm{csc}\left(B\right)\mathrm{tan}\left(x\right)$.

Rewrite $\mathrm{csc}\left(B\right)$ in terms of sines and cosines.

$\mathrm{cos}\left(B\right)\frac{1}{\mathrm{sin}\left(B\right)}\mathrm{tan}\left(x\right)=1$

Combine $\mathrm{cos}\left(B\right)$ and $\frac{1}{\mathrm{sin}\left(B\right)}$.

$\frac{\mathrm{cos}\left(B\right)}{\mathrm{sin}\left(B\right)}\mathrm{tan}\left(x\right)=1$

Rewrite $\mathrm{tan}\left(x\right)$ in terms of sines and cosines.

$\frac{\mathrm{cos}\left(B\right)}{\mathrm{sin}\left(B\right)}\cdot \frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}=1$

Multiply $\frac{\mathrm{cos}\left(B\right)}{\mathrm{sin}\left(B\right)}$ by $\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$.

$\frac{\mathrm{cos}\left(B\right)\mathrm{sin}\left(x\right)}{\mathrm{sin}\left(B\right)\mathrm{cos}\left(x\right)}=1$

Separate fractions.

$\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}\cdot \frac{\mathrm{cos}\left(B\right)}{\mathrm{sin}\left(B\right)}=1$

Convert from $\frac{\mathrm{cos}\left(B\right)}{\mathrm{sin}\left(B\right)}$ to $\mathrm{cot}\left(B\right)$.

$\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}\mathrm{cot}\left(B\right)=1$

$\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$ to $\mathrm{tan}\left(x\right)$.

$\mathrm{tan}\left(x\right)\mathrm{cot}\left(B\right)=1$

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