Question #0cc73

Apr 10, 2017

${\sec}^{2} \left(x\right) \cot \left(x\right) - \cot \left(x\right) = \tan \left(x\right)$

Explanation:

$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

and,

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

expanding the given function;

$\frac{1}{\cos} ^ 2 \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right) - \cos \frac{x}{\sin} \left(x\right)$

$\implies \frac{1}{\cos \left(x\right) \cdot \sin \left(x\right)} - \cos \frac{x}{\sin} \left(x\right)$

Multiplying the numerator and denominator of $\cos \frac{x}{\sin} \left(x\right)$ by $\cos \left(x\right)$ we get the same denominators for both terms.

$\implies \frac{1}{\cos \left(x\right) \cdot \sin \left(x\right)} - {\cos}^{2} \frac{x}{\cos \left(x\right) \cdot \sin \left(x\right)}$

Then we can simplify further by taking common denominator;

$\implies \frac{1 - {\cos}^{2} \left(x\right)}{\cos \left(x\right) \cdot \sin \left(x\right)}$

we know, that $1 - {\cos}^{2} \left(x\right) = {\sin}^{2} \left(x\right)$

Therefore, we have

${\sin}^{2} \frac{x}{\cos \left(x\right) \cdot \sin \left(x\right)}$

$\implies {\sin}^{\cancel{2}} \frac{x}{\cos \left(x\right) \cdot \cancel{\sin \left(x\right)}}$

$\implies \sin \frac{x}{\cos} \left(x\right)$

$\implies \tan \left(x\right)$