# TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES

Two acute angles are complementary to each other if their sum is equal to 90°. In a right triangle the sum of the two acute angles is equal to 90°. So, the two acute angles of a right triangle are always complementary to each other.

Let ABC be a right triangle, right angled at B.

If  ACB = θ, then  BAC = 90° - θ and hence the angles  BAC and  ACB are complementary

For the angle θ, we have

 sin θ =  AB/ACcos θ  =  BC/ACtan θ  =  AB/BC cosec θ =  AC/ABsec θ  =  AC/BCcot θ  =  BC/AB

Similarly, for the angle (90° - θ), we have

 cos (90 - θ) =  AB/ACsin (90 - θ)  =  BC/ACtan (90 - θ)  =  BC/AB cosec (90 - θ)  =  AC/ABsec (90 - θ)  =  AC/BCcot (90 - θ)  =  AB/BC

Comparing the equations in (1) and (2) we get,

sin θ  =  AB/AC  =  cos (90 - θ)

cos θ  =  BC/AC  =  sin (90 - θ)

tan θ  =  AB/BC  =  cot (90 - θ)

cosec θ  =  AC/AB  =  sec (90 - θ)

sec θ  =  AC/BC  =  cosec (90 - θ)

cot θ  =  BC/AB  =  tan (90 - θ)

## Trigonometric Ratios of Complementary Angles

 sin θ  =  cos (90 - θ)cos θ  =  sin (90 - θ)tan θ  =  cot (90 - θ) cosec θ  =  sec (90 - θ)sec θ  =  cosec (90 - θ)cot θ  =  tan (90 - θ)

Question 1 :

Evaluate

cos 56° / sin 34°

The angles 56° and 34° are complementary.

Use trigonometric ratios of complementary angles.

cos 56° / sin 34°  =  cos 56° / cos (90° - 34°)

cos 56° / sin 34°  =  cos 56° / cos 56°

cos 56° / sin 34°  =  1

Question 2 :

Evaluate

tan 25° / cot 65°

The angles 25° and 65° are complementary.

Use trigonometric ratios of complementary angles.

tan 25° / cot 65°  =  tan 25) / tan (90° - 65°)

tan 25° / cot 65°  =  tan 25° / tan 25°

tan 25° / cot 65°  =  1

Question 3 :

Evaluate

(cos 65° sin 18° cos 58°)/(cos 72° sin 25° sin 32°)

Use trigonometric ratios of complementary angles.

cos 65°  =  cos (90° - 25°)  =  sin 25°

sin 18°  =  sin (90° - 72°)  =  cos 72°

cos 58°  =  cos (90° - 32°)  =  sin 32°

(cos 65° sin 18° cos 58°) / (cos 72° sin 25° sin 32°) is

=  (sin 25° cos 72° sin 32°) / (cos 72° sin 25° sin 32°)

=  1

So,

cos 65° sin 18° cos 58°)/(cos 72° sin 25° sin 32°  =  1

Question 4 :

Evaluate :

tan 35° tan 60° tan 55° tan 30°

tan 35°  =  tan (90° - 55°)  =  cot 55°  =  1/tan 55°

tan 60°  =  tan (90° - 30°)  =  cot 30°  =  1/tan 30°

tan 35° tan 60° tan 55° tan 30° :

=  (1/tan 55° ) x (1/tan 30°) tan 55° tan 30°

=  1

So,

tan 35° tan 60° tan 55° tan 30°   =  1

Question 5 :

If sin A  =  cos 33°, find A.

sin A  =  cos 33°

sin A  =  sin (90° - 33°)

sin A  =  sin 57°

A  =  57°

Question 6 :

If tan A tan 35°  =  1, find A.

tan A tan 35°  =  1

Divide  each side by tan 35°.

tan A  =  1/tan 35°

tan A  =  cot 35°

tan A  =  tan (90° - 35°)

tan A  =  tan 55°

A  =  55°

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