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7. Trigonometric ratios of complementary angles

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