Trigonometric Ratios Of Complementary Angles

Let us construct a right-angled ΔABC in which ∠ABC = θ, ∠BCA = 90º.

Then we have ∠BAC = (90º – θ). Angles θ and (90º – θ) are defined as complementary angles.

In figure (i), the sides are labelled corresponding to angle θ and in figure (ii), the sides are labelled corresponding to angle (90º – θ).

From figure (ii), we have sin (90º – θ)= BC/BA and from figure (i), we have BC/BA = cos θ.

∴ sin (90º – θ)= BC/BA = cos θ.

Similarly, cos (90º – θ)= AC/BA = sIN θ

tan (90º – θ)=

cosec (90º – θ)=

sec (90º – θ) =

cot (90º – θ) = .

If A and B are two complementary acute angles, i.e., A + B = 90º, then we have

sin A = sin (90º – B) = cos B

cos A = cos (90º – B) = sin B

tan A = tan (90º – B) = cot B

cosec A = cosec (90º – B) = sec B

sec A = sec (90º – B) = cosec B

cot A = cot (90º – B) = tan B.