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

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$\frac{\tan\left(x\right)}{\sec\left(x\right)-1}$

Learn how to solve trigonometric identities problems step by step online. Prove the trigonometric identity (sec(x)+1)/tan(x)=tan(x)/(sec(x)-1). Starting from the right-hand side (RHS) of the identity. Multiply and divide the fraction \frac{\tan\left(x\right)}{\sec\left(x\right)-1} by the conjugate of it's denominator \sec\left(x\right)-1. Multiplying fractions \frac{\tan\left(x\right)}{\sec\left(x\right)-1} \times \frac{\sec\left(x\right)+1}{\sec\left(x\right)+1}. 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..