### Theory:

Let us recall that if the sum of the two acute angles is $$90^{\circ}$$, then the angles are said to be complementary.

In a right-angled triangle, the sum of two acute angles are $$90^{\circ}$$.

That is, we can say that the two acute angles in a right-angled triangle are complementary.

Consider the triangle $$PQR$$ right-angled at $$P$$. Here $$R$$ and $$Q$$ are complementary angles.

Therefore if $$\angle R$$ $$=$$ $$\theta$$, then $$\angle Q$$ $$=$$ $$90^{\circ} - \theta$$.

Let us write all the trigonometric ratios with respect to $$\angle R$$ $$=$$ $$\theta$$ in a table.

Table $$1$$:

 Trigonometric ratio Relationship with the $$\Delta PQR$$ Trigonometric ratio Relationship with the $$\Delta PQR$$ $$\sin \theta$$ $$\sin \theta$$ $$=$$ $$\frac{PQ}{RQ}$$ $$\text{cosec}\,\theta$$ $$\text{cosec}\,\theta$$ $$=$$ $$\frac{RQ}{PQ}$$ $$\cos \theta$$ $$\cos \theta$$ $$=$$ $$\frac{PR}{RQ}$$ $$\sec \theta$$ $$\sec \theta$$ $$=$$ $$\frac{RQ}{PR}$$ $$\tan \theta$$ $$\tan \theta$$ $$=$$ $$\frac{PQ}{PR}$$ $$\cot \theta$$ $$\cot \theta$$ $$=$$ $$\frac{PR}{PQ}$$

Now let us write all the trigonometric ratios with respect to $$\angle Q$$ $$=$$ $$90^{\circ} - \theta$$ in a table.

Table $$2$$:

 Trigonometric ratio Relationship with the $$\Delta PQR$$ Trigonometric ratio Relationship with the $$\Delta PQR$$ $$\sin (90^{\circ} - \theta)$$ $$\sin (90^{\circ} - \theta)$$ $$=$$ $$\frac{PR}{RQ}$$ $$\text{cosec}\,(90^{\circ} - \theta)$$ $$\text{cosec}\,(90^{\circ} - \theta)$$ $$=$$ $$\frac{RQ}{PR}$$ $$\cos (90^{\circ} - \theta)$$ $$\cos (90^{\circ} - \theta)$$ $$=$$ $$\frac{PQ}{RQ}$$ $$\sec (90^{\circ} - \theta)$$ $$\sec (90^{\circ} - \theta)$$ $$=$$ $$\frac{RQ}{PQ}$$ $$\tan (90^{\circ} - \theta)$$ $$\tan (90^{\circ} - \theta)$$ $$=$$ $$\frac{PR}{PQ}$$ $$\cot (90^{\circ} - \theta)$$ $$\cot (90^{\circ} - \theta)$$ $$=$$ $$\frac{PQ}{PR}$$

Comparing the tables $$1$$ and $$2$$, we arrive at the following identities.

1. $$\sin \theta$$ $$=$$ $$\cos (90^{\circ} - \theta)$$

2. $$\cos \theta$$ $$=$$ $$\sin (90^{\circ} - \theta)$$

3. $$\tan \theta$$ $$=$$ $$\cot (90^{\circ} - \theta)$$

4. $$\text{cosec}\,\theta$$ $$=$$ $$\sec (90^{\circ} - \theta)$$

5. $$\sec \theta$$ $$=$$ $$\text{cosec}(90^{\circ} - \theta)$$

6. $$\cot \theta$$ $$=$$ $$\tan (90^{\circ} - \theta)$$