Trigonometric Ratios Of Complementary Angles

Trigonometric Ratios Of Complementary Angles

We know 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:    [latex s=2] \text{Evaluate }\frac{\tan 65{}^\circ }{\cot 25{}^\circ }.  [/latex]
Sol.    ∵ 65° + 25° = 90°
[latex s=2] \therefore \frac{\tan 65{}^\circ }{\cot 25{}^\circ }=\frac{\tan \,(90{}^\circ -25{}^\circ )}{\cot \,25{}^\circ }=\frac{\cot 25{}^\circ }{\cot 25{}^\circ }=1 [/latex]

Example 2:    Without using trigonometric tables, evaluate the following:
[latex s=2]\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}17\acute{\ }}{\cos \,\,59{}^\text{o}\,43\acute{\ }}[/latex]
Sol.    (i)  We have,
[latex s=2] \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 [/latex]
[∵  cos(90º – θ) = sin θ]
(ii)  We have,
[latex s=2] \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 [/latex]
[∵  sin (90º – θ) = cos θ]
(iii)  We have,
[latex s=2] \frac{\sin \,\,30{}^\text{o}\,17\acute{\ }}{\cos \,\,59{}^\text{o}\,43\acute{\ }}=\frac{\sin (90{}^\text{o}-59{}^\text{o}43\acute{\ })}{\cos \,59{}^\text{o}43\acute{\ }}=\frac{\cos \,59{}^\text{o}43\acute{\ }}{\cos \,59{}^\text{o}43\acute{\ }}=1 [/latex]

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:    [latex s=2] (\text{i})\text{ }\frac{\cot \,\,54{}^\text{o}}{\tan \,\,36{}^\text{o}}+\frac{\tan \,\,20{}^\text{o}}{\cot \,\,70{}^\text{o}}-2 [/latex]
(ii)  sec 50º sin 40° + cos 40º cosec 50º
Sol.   (i)  We have,
[latex s=2] \frac{\cot \,\,54{}^\text{o}}{\tan \,\,36{}^\text{o}}+\frac{\tan \,\,20{}^\text{o}}{\cot \,\,70{}^\text{o}}-2 [/latex]
[latex s=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 [/latex]
[latex s=2] =\frac{\tan \,\,36{}^\text{o}}{\tan \,\,36{}^\text{o}}+\frac{\tan \,\,20{}^\text{o}}{\tan \,\,20{}^\text{o}}-2[/latex]
= 1 + 1 – 2 = 0
(ii)  We have,
sec50º sin40º + cos40º cosec50º
= sec(90º – 40º) sin40º + cos40º cosec(90º – 40º)
= cosec40º sin40º + cos40ºsec40º
[latex s=2] =\frac{\sin \,\,40{}^\text{o}}{\sin \,\,40{}^\text{o}}+\frac{\cos \,\,40{}^\text{o}}{\cos \,\,40{}^\text{o}}[/latex]
= 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:
[latex s=2] \frac{{{\sin }^{2}}20{}^\text{o}+{{\sin }^{2}}70{}^\text{o}}{{{\cos }^{2}}20{}^\text{o}+{{\cos }^{2}}70{}^\text{o}}+\frac{\sin (90{}^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90{}^\text{o}-\theta )\cos \theta }{\cot \theta } [/latex]
Sol.       [latex s=2] \frac{{{\sin }^{2}}20{}^\text{o}+{{\sin }^{2}}70{}^\text{o}}{{{\cos }^{2}}20{}^\text{o}+{{\cos }^{2}}70{}^\text{o}}+\frac{\sin (90{}^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90{}^\text{o}-\theta )\cos \theta }{\cot \theta } [/latex]
[latex s=2] =\frac{{{\sin }^{2}}20{}^\text{o}+{{\sin }^{2}}(90{}^\text{o}-20{}^\text{o})}{{{\cos }^{2}}20{}^\text{o}+{{\cos }^{2}}(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 } [/latex]
[latex s=2] =\frac{{{\sin }^{2}}20{}^\text{o}+{{\cos }^{2}}20{}^\text{o}}{{{\cos }^{2}}20{}^\text{o}+{{\sin }^{2}}20{}^\text{o}}+\frac{\cos \theta \sin \theta }{\frac{\sin \theta }{\cos \theta }}+\frac{\sin \theta \cos \theta }{\frac{\cos \theta }{\sin \theta }} [/latex]
[latex] \left[ \sin (90{}^\text{o}-\theta )=\cos \theta \,\,\,and\cos (90{}^\text{o}-\theta )\,\,=\,\,\sin \theta \right] [/latex]
= 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
[latex s=2] \tan \frac{B+C}{2}=\cot \frac{A}{2} [/latex]
Sol.    In ∆ABC, we have
A + B + C = 180º
⇒ B + C = 180º – A
[latex s=2] \Rightarrow \frac{B+C}{2}=\text{ }90{}^\text{o}-\frac{A}{2} [/latex]
[latex s=2] \Rightarrow \tan \left( \frac{B+C}{2} \right)=\tan \left( 90{}^\text{o}-\frac{A}{2} \right) [/latex]
[latex s=2] \Rightarrow \tan \left( \frac{B+C}{2} \right)=\cot \frac{A}{2} [/latex]

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
[latex s=2] \sin \left( \frac{B+C}{2} \right)=\cos \frac{A}{2}  [/latex]
Sol.    ∵  A + B + C = 180° (a.s.p. of ∆)
B + C = 180° – A
[latex s=2]\left( \frac{B+C}{2} \right)=90{}^\circ -\frac{A}{2}[/latex]
[latex s=2] \sin \left( \frac{B+C}{2} \right)=\sin \left( 90{}^\circ -\frac{A}{2} \right) [/latex]
[latex s=2] \sin \left( \frac{B+C}{2} \right)=\cos \frac{A}{2}  [/latex]      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°.