Q.Prove that $\dfrac{\tan\theta}{\sec\theta − 1} = \dfrac{\sec\theta + 1}{\tan\theta}$.

A trigonometric identity is an equation in trigonometric functions that holds for every admissible value of the angle, not just for special ones. The three fundamental identities all descend from a single geometric fact — that a point (cosθ, sinθ) always lies on the unit circle, so $x^2+y^2=1$ becomes

Dividing this identity through by $\cos^2\theta$ gives $\tan^2\theta + 1 = \sec^2\theta$, i.e. $1+\tan^2\theta=\sec^2\theta$ (valid wherever cosθ ≠ 0); dividing instead by $\sin^2\theta$ gives $1+\cot^2\theta=\text{cosec}^2\theta$ (valid wherever sinθ ≠ 0). These three identities are the basic toolkit for simplifying trigonometric expressions: they let you rewrite $\sec^2\theta - 1$ as $\tan^2\theta$, combine fractions with denominators like $1-\sin\theta$ and $1+\sin\theta$ (whose product is $\cos^2\theta$), find one tri …