# tan(x)sin(x) + sec(x)cos^2(x) = sec(x)

$$\displaystyle{\tan{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}+{\sec{{\left({x}\right)}}}{{\cos}^{{2}}{\left({x}\right)}}={\sec{{\left({x}\right)}}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Nicole Conner

$$\displaystyle{\tan{{\left({x}\right)}}}=\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}.$$
$$\displaystyle{\tan{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}=\frac{{{\sin}^{{2}}{\left({x}\right)}}}{{\cos{{\left({x}\right)}}}}.$$
$$\displaystyle{\sec{{\left({x}\right)}}}=\frac{{1}}{{\cos{{\left({x}\right)}}}};$$
$$\displaystyle{\sec{{\left({x}\right)}}}{{\cos}^{{2}}{\left({x}\right)}}={\cos{{\left({x}\right)}}}.$$
$$\tan(x)\sin(x) + \sec(x)\cos^2(x) = \frac{\sin^2(x)}{\cos(x)} + \cos(x)$$

$$= \frac{\sin^2(x) + \cos^2(x)}{\cos(x)} = \frac{1}{\cos(x)} = \sec(x) \{proved\}$$
as we know $$\displaystyle{{\sin}^{{2}}{\left({x}\right)}}+{{\cos}^{{2}}{\left({x}\right)}}={1}$$.