Math Instructor

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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