Prove the trigonometric identity $\frac{1}{\sec\left(x\right)+\tan\left(x\right)}=\sec\left(x\right)-\tan\left(x\right)$

Learn how to solve factor by difference of squares problems step by step online. Prove the trigonometric identity 1/(sec(x)+tan(x))=sec(x)-tan(x). Starting from the left-hand side (LHS) of the identity. Multiply and divide the fraction \frac{1}{\sec\left(x\right)+\tan\left(x\right)} by the conjugate of it's denominator \sec\left(x\right)+\tan\left(x\right). Multiplying fractions \frac{1}{\sec\left(x\right)+\tan\left(x\right)} \times \frac{\sec\left(x\right)-\tan\left(x\right)}{\sec\left(x\right)-\tan\left(x\right)}. The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: (a+b)(a-b)=a^2-b^2..