Q.If $y = \dfrac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}$, then prove that $\dfrac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=y$.

The six trigonometric ratios (right-triangle definition). For an acute angle $\theta$ in a right triangle, with sides labelled relative to $\theta$ as opposite, adjacent, and the hypotenuse:

The other four ratios are always built from these two:

Only $\sin\theta$ and $\cos\theta$ are truly independent — every other ratio is a quotient or reciprocal built from them, which is why rewriting a hard trig expression purely in $\sin,\cos$ is such a reliable first move.

Exact values at standard angles ($0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$):

$\tan90^\circ,\sec90^\circ$ are undefined because $\cos90^\circ=0$; $\csc0^\circ,\cot0^\circ$ are undefined because $\sin0^\circ=0$. Also $\sin30^\circ=\cos60^\circ$ and $\sin60^\circ=\cos30^\circ$ — an early instance of the general complementary-angle pattern $\sin\theta=\cos(90^\circ-\theta)$.

What makes an equation an identity. A trigonometric identity is an equation in trigonometric ratios holding for every value of $\theta$ in its domain — not merely for some particular angle. $\sec\theta=\dfrac{1}{\cos\theta}$ is an identity (true for all $\theta$ with $\cos\theta\ne0$); $\sin\theta=\dfrac12$ is not (true only at specific angles like $30^\circ$ or $150^\circ$).

The three fundamental (Pythagorean) identities, obtained from the Pythagorean theorem applied to a right triangle, divided in turn by the square of the hypotenuse, the adjacent side, and the opposite side:

Here $\sin^2\theta$ means $(\sin\theta)^2$, and similarly for the other ratios. Each identity holds wherever both sides are defined — e.g. $\sec^2\theta-\tan^2\theta=1$ says nothing at $\theta=90^\circ$, where both terms are individually undefined, but this doesn't stop it from being a genuine identity for every $\theta$ where it does make sense.

Reciprocal and quotient identities (restating the ratio definitions as identities in their own right): …