# How do you prove  [(1)/(1-sinx)]+[(1)/(1+sinx)]=2sec^2x?

##### 1 Answer
Apr 23, 2015

The formula can be proven by applying: 1) Least common multiple; 2) applying the trigonometric entity ${\sin}^{2} x + {\cos}^{2} x = 1$

#### Explanation:

Head

Key-relation : ${\sin}^{2} x + {\cos}^{2} x = 1$

Key-concept: Least common multiple; when no common multiples, just multiply the terms in the denominator.

Calculation

The above formula can be proven by transforming left side to right side:

$\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = \frac{1 + \sin x + 1 - \sin x}{\left(1 + \sin x\right) \left(1 - \sin x\right)}$

To arrive to right-hand side, just divide the denominator to $\left(1 + \sin x\right) \left(1 - \sin x\right)$, the least common multiple, and multiply the numerator to the remaining, since they are all 1, just put the value.

By simple algebra and make use of $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$, it can be seen from normal multiplication.

$\frac{1 + \sin x + 1 - \sin x}{\left(1 + \sin x\right) \left(1 - \sin x\right)} = \frac{2}{1 - {\sin}^{2} x}$

Finally apply: ${\sin}^{2} x + {\cos}^{2} x = 1$, which gives out ${\cos}^{2} x = 1 - {\sin}^{2} x$

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

To finish, remember that $\sec x = \frac{1}{\cos} x$, hence:

$2 \cdot {\left(\frac{1}{\cos} x\right)}^{2} = 2 {\sec}^{2} x$