# TRIGONOMETRIC RATIOS FOR COMPLEMENTARY ANGLES EXAMPLES

## About "Trigonometric Ratios for Complementary Angles Examples"

Trigonometric Ratios for Complementary Angles Examples :

Here we are going to see some example problems on trigonometric ratios for complementary angles.

Two acute angles are said to be complementary if the sum of their measures is equal to 90° .

sin θ = cos (90° - θ)  <==> cos θ = sin (90° - θ

cosec θ = sec (90° - θ)  <==> sec θ = cosec (90° - θ

tan θ = cot (90° - θ)  <==> cot θ = tan (90° - θ)

## Trigonometric Ratios for Complementary Angles Examples - Practice questions

Question 1 :

Find the value of the following:

(i)  (cos 47/sin 43)2 + (sin 72/cos 18)2 - 2cos2 45

Solution :

=  (cos 47/sin 43)+ (sin 72/cos 18)2 - 2cos2 45

cos 47  = sin (90 - 47)  =  sin 43

sin 72  =  cos (90 - 72)  =  cos 18

cos 45  =  1/√2

=  (sin 43/sin 43)+ (cos 18/cos 18)2 - 2 (1/√2)2

=  1+ 12 - 1

=  1

(ii)  (cos 70/sin 20)+ (cos 59/sin 31)2 + cos θ/sin(90-θ) - 8 cos2 60

L.H.S :

=  (cos 70/sin 20)+ (cos 59/sin 31)2 + cos θ/sin(90-θ) - 8 cos2 60

cos 70  =  sin (90 - 70)  =  sin 20

cos 59  =  sin (90 - 59)  =  sin 31

sin(90-θ)  =  cos θ

cos2 60  =  (1/2)=  1/4

=  (sin 20/sin 20)+ (sin 31/sin 31)2 + (cos θ/cosθ) - 8(1/4)

=  1 + 1 + 1 - 2

=  3 - 2

=  1

(iii) tan 15° tan 30° tan 45° tan 60° tan 75°

Solution :

=  tan 15° tan 30° tan 45° tan 60° tan 75°

tan 75°  =  cot (90 - 15)  =  cot 15

tan 30°  =  1/√3, tan 45°  =  1, tan 60°  =  √3

=  tan 15° (1/√3) 1 (√3) cot 15°

=  tan 15° (1/tan 15°)

=  1

(iv)  [cot θ / tan (90 - θ)] + [cos (90 - θ) tan θ sec(90 - θ)/sin (90 - θ) cot (90 - θ) cosec (90 - θ)]

Solution :

=  [cot θ / tan (90 - θ)] + [cos (90 - θ) tan θ sec(90 - θ)/sin (90 - θ) cot (90 - θ) cosec (90 - θ)]

=  [cot θ/cot θ+ [sin θ tan θ cosec θ/cos θ tan θ sec θ]

=  1 + (sin θ cosec θ/cos θ sec θ)

=  1 + 1

=  2

After having gone through the stuff given above, we hope that the students would have understood, "Trigonometric Ratios for Complementary Angles Examples"

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