Trigonometric Ratios Of Complementary Angles
We know Trigonometric ratios of complementary angles are pair of angles whose sum is 90°
Like 40°, 50°, 60°, 30°, 20°, 70°, 15°, 75° ; etc,
Formulae:
sin (90° – θ) = cos θ, cot (90° – θ) = tanθ
cos (90° – θ) = sin θ, sec (90° – θ) = cosec θ
tan (90° – θ) = cot θ, cosec (90° – θ) = sec θ
Trigonometric Ratios Of Complementary Angles With Examples
Example 1: ( text{Evaluate }frac{tan 65{}^circ }{cot 25{}^circ }. )
Sol. ∵ 65° + 25° = 90°
( frac{tan 65{}^circ }{cot 25{}^circ }=frac{tan ,(90{}^circ -25{}^circ )}{cot ,25{}^circ }=frac{cot 25{}^circ }{cot 25{}^circ }=1 )
Example 2: Without using trigonometric tables, evaluate the following:
(left( i right)~text{ }frac{cos ,,37{}^text{o}}{sin ,,53{}^text{o}}text{ }left( ii right)~frac{sin ,,41{}^text{o}}{cos ,,49{}^text{o}}~text{ }left( iii right)~frac{sin ,,30{}^text{o}17acute{ }}{cos ,,59{}^text{o},43acute{ }})
Sol. (i) We have,
( frac{cos ,,37{}^text{o}}{sin ,,53{}^text{o}}=frac{cos (90{}^text{o}-53{}^text{o})}{sin ,,53{}^text{o}}=frac{sin ,,53{}^text{o}}{sin ,,53{}^text{o}}=1 )
[∵ cos(90º – θ) = sin θ]
(ii) We have,
( frac{sin ,,41{}^text{o}}{cos ,,49{}^text{o}}=frac{sin (90{}^text{o}-49{}^text{o})}{cos ,49{}^text{o}}=frac{cos ,49{}^text{o}}{cos ,49{}^text{o}}=1 )
[∵ sin (90º – θ) = cos θ]
(iii) We have,
( frac{sin ,,30{}^text{o},17acute{ }}{cos ,,59{}^text{o},43acute{ }}=frac{sin (90{}^text{o}-59{}^text{o}43acute{ })}{cos ,59{}^text{o}43acute{ }}=frac{cos ,59{}^text{o}43acute{ }}{cos ,59{}^text{o}43acute{ }}=1 )
[youtube https://www.youtube.com/watch?v=jo6-yeD4YBM?start=59&feature=oembed]
Example 3: Without using trigonometric tables evaluate the following:
(i) sin2 25º + sin2 65º (ii) cos2 13º – sin2 77º
Sol. (i) We have,
sin2 25º + sin265º = sin2 (90º – 65º) + sin2 65º
= cos265º + sin265º = 1
[∵ sin (90º – θ) = cos θ]
(ii) We have,
cos213º– sin277º = cos2(90º – 77º) – sin277º
= sin277º – sin277º = 0
[∵ cos (90º – θ) = sin θ]
Example 4: ( (text{i})text{ }frac{cot ,,54{}^text{o}}{tan ,,36{}^text{o}}+frac{tan ,,20{}^text{o}}{cot ,,70{}^text{o}}-2 )
(ii) sec 50º sin 40° + cos 40º cosec 50º
Sol. (i) We have,
( frac{cot ,,54{}^text{o}}{tan ,,36{}^text{o}}+frac{tan ,,20{}^text{o}}{cot ,,70{}^text{o}}-2 )
( =frac{cot (90{}^text{o}-36{}^text{o})}{tan 36{}^text{o}}+frac{tan ,20{}^text{o}}{cot (90{}^text{o}-20{}^text{o})}-2 )
( =frac{tan ,,36{}^text{o}}{tan ,,36{}^text{o}}+frac{tan ,,20{}^text{o}}{tan ,,20{}^text{o}}-2)
= 1 + 1 – 2 = 0
(ii) We have,
sec50º sin40º + cos40º cosec50º
= sec(90º – 40º) sin40º + cos40º cosec(90º – 40º)
= cosec40º sin40º + cos40ºsec40º
( =frac{sin ,,40{}^text{o}}{sin ,,40{}^text{o}}+frac{cos ,,40{}^text{o}}{cos ,,40{}^text{o}})
= 1 + 1 = 2
Example 5: Express each of the following in terms of trigonometric ratios of angles between 0º and 45º;
(i) cosec 69º + cot 69º
(ii) sin 81º + tan 81º
(iii) sin 72º + cot 72º
Sol. (i) We have,
cosec 69º + cot 69º
= cosec (90º – 21º) + cot (90º – 21º)
= sec 21º + tan 21º
[∵ cosec (90º – θ) = sec θ and cot (90º –θ) = tan θ]
(ii) We have,
sin 81º + tan 81º
= sin (90º – 9º) + tan (90º – 9º)
= cos 9º + cot 9º
[∵ sin (90º – θ) = cos θ and tan (90º –θ) = cot θ]
(iii) We have,
sin 72º + cot 72º
= sin (90º – 18º) + cot (90º – 18º)
= cos 18º + tan 18º
[∵ sin (90º – 18º) = cos 18º and tan (90º –18º) = cot 18º]
Example :6 Without using trigonometric tables, evaluate the following:
( frac20{}^text{o}+70{}^text{o}}20{}^text{o}+70{}^text{o}}+frac{sin (90{}^text{o}-theta )sin theta }{tan theta }+frac{cos (90{}^text{o}-theta )cos theta }{cot theta } )
Sol. ( frac20{}^text{o}+70{}^text{o}}20{}^text{o}+70{}^text{o}}+frac{sin (90{}^text{o}-theta )sin theta }{tan theta }+frac{cos (90{}^text{o}-theta )cos theta }{cot theta } )
( =frac20{}^text{o}+(90{}^text{o}-20{}^text{o})}20{}^text{o}+(90{}^text{o}-20{}^text{o})}+frac{sin (90{}^text{o}-theta )sin theta }{tan theta }+frac{cos (90{}^text{o}-theta )cos theta }{cot theta } )
( =frac20{}^text{o}+20{}^text{o}}20{}^text{o}+20{}^text{o}}+frac{cos theta sin theta }{frac{sin theta }{cos theta }}+frac{sin theta cos theta }{frac{cos theta }{sin theta }} )
( left[ sin (90{}^text{o}-theta )=cos theta ,,,andcos (90{}^text{o}-theta ),,=,,sin theta right] )
= 1 + cos2 θ + sin2 θ = 1 + 1 = 2
Example :7 If tan 2θ = cot (θ + 6º), where 2θ and θ + 6º are acute angles, find the value of θ.
Sol. We have,
tan 2θ = cot (θ + 6º)
⇒ cot(90º – 2θ) = cot (θ + 6º)
⇒ 90º – 2θ = θ + 6º ⇒ 3θ = 84º
⇒ θ = 28º
Example :8 If A, B, C are the interior angles of a triangle ABC, prove that
( tan frac{B+C}{2}=cot frac{A}{2} )
Sol. In ∆ABC, we have
A + B + C = 180º
⇒ B + C = 180º – A
( Rightarrow frac{B+C}{2}=text{ }90{}^text{o}-frac{A}{2} )
( Rightarrow tan left( frac{B+C}{2} right)=tan left( 90{}^text{o}-frac{A}{2} right) )
( Rightarrow tan left( frac{B+C}{2} right)=cot frac{A}{2} )
Example :9 If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Sol. tan 2A = cot (A – 18°)
cot (90° – 2A) = cot (A – 18°)
(∵ cot (90° – θ) = tan θ)
90° – 2A = A – 18°
3A = 108°
A = 36°
Example :10 If tan A = cot B, prove that A + B = 90°.
Sol. ∵ tan A = cot B
tan A = tan (90° – B)
A = 90° – B
A + B = 90°. Proved
Example :11 If A, B and C are interior angles of a triangle ABC, then show that
( sin left( frac{B+C}{2} right)=cos frac{A}{2} )
Sol. ∵ A + B + C = 180° (a.s.p. of ∆)
B + C = 180° – A
(left( frac{B+C}{2} right)=90{}^circ -frac{A}{2})
( sin left( frac{B+C}{2} right)=sin left( 90{}^circ -frac{A}{2} right) )
( sin left( frac{B+C}{2} right)=cos frac{A}{2} ) Proved.
Example :12 Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Sol. ∵ 23 = 90 – 67 & 15 = 90 – 75
∴ sin 67° + cos 75°
= sin (90 – 23)° + cos (90 – 15)°
= cos 23° + sin 15°.
