Trigonometric Ratios of Complementary Angles

We know that angles are said to be complementary if their sum is equal to 90$^\circ$.

For example, we have $\triangle ABC$

We know that in $\triangle ABC$, $\angle A$ + $\angle B$ + $\angle C$ = 180$^\circ$

We have $\angle C =90^\circ$

$\Rightarrow \angle A + \angle B = 180-90 = 90^\circ$

Therefore, $\angle A$ and $\angle B$ are complementary angles.

Lets find out relation between them.

$sin A = \frac{BC}{AB}, cos A =\frac{AC}{AB}, tan A =\frac{BC}{AC}$

$cosec A =\frac{AB}{BC}, sec A =\frac{AB}{AC}, cot A =\frac{AC}{BC}$

$sin B = sin (90^\circ-A) = \frac{AC}{AB}$

$cos B = cos(90^\circ-A) = \frac{BC}{AB}$

$tan B = tan(90^\circ-A)=\frac{AC}{BC}$

$cosec B = cosec (90^\circ-A) = \frac{AB}{AC}$

$sec B = sec(90^\circ-A) = \frac{AB}{BC}$

$cot B = cot (90^\circ-A)= \frac{BC}{AC}$

From above information we can say that

$sin A = cos(90^\circ-A)$

$sin (90^\circ-A) = cos A$

$tan A = cot(90^\circ-A)$

$cot A = tan (90^\circ-A)$

$sec A = cosec(90^\circ-A)$

$cosec A = sec(90^\circ-A)$

Posted in : NCERT Solutions, Trigonometry

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