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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