Question

Verify this identity starting from the left

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

• sin²x + cos²x = 1

• cot x = $\frac{cosx}{sinx}$, csc x = $\frac{1}{sinx}$

Consider the left side

$\frac{1+cosx}{1-cosx}$ - $\frac{1-cosx}{1+cosx}$

Expressing as a single fraction

= $\frac{(1+cosx)^2-(1-cosx)^2}{(1-cosx)(1+cosx)}$

Expand and simplify numerator/ denominator

= $\frac{1+2cosx+cos^2x-1+2cosx-cos^2x}{1-cos^2x}$

= $\frac{4cosx}{sin^2x}$

= $\frac{4cosx}{sinx}$ × $\frac{1}{sinx}$

= 4cotxcscx = right side ⇒ verified