How to find the trigonometrical ratios of complementary angles?
If the sum of two angles is one right angle or 90°, then one angle is said to be complementary of the other. Thus, 25° and 65°; θ° and (90 - θ)° are complementary to each other.
Suppose a rotating line rotates about O in the anti-clockwise sense and starting from its initial position
\(\overrightarrow{OX}\) traces out angle ∠XOY = θ, where θ is acute.
Take a point P on \(\overrightarrow{OY}\) and draw \(\overline{PQ}\) perpendicular to OX. Let, ∠OPQ = α. Then, we have,
α + θ = 90°
or, α = 90° - θ.
Therefore, θ and α are complementary to each other.
Now, by the definition of trigonometric ratio,
sin θ = \(\frac{\overline{PQ}}{\overline{OP}}\); ………. (i)
cos θ = \(\frac{\overline{OQ}}{\overline{OP}}\); ………. (ii)
tan θ = \(\frac{\overline{PQ}}{\overline{OQ}}\) ………. (iii)
And sin α = \(\frac{\overline{OQ}}{\overline{OP}}\); ………. (iv)
cos α = \(\frac{\overline{PQ}}{\overline{OP}}\); ………. (v)
tan α = \(\frac{\overline{OQ}}{\overline{PQ}}\) ….… (vi)
From (i) and (iv) we have,
sin α = cos θ
or, sin (90° - θ) = cos θ;
From (ii) and (v) we have,
cos α = sin θ
or, cos (90° - θ) = sin θ;
From (iii) and (vi) we have,
And tan α = 1/tan θ
or, tan (90° - θ) = cot θ.
Similarly, csc (90° - θ) = sec θ;
sec (90° - θ) = csc θ
and cot (90° - θ) = tan θ.
Therefore,
Sine of any angle = cosine of its complementary angle;
Cosine of any angle = sine of its complementary angle;
Tangent of any angle = cotangent of its complementary angle.
Corollary:
Complementary Angles: Two angles are said to be complementary if their sum is 90°. Thus θ and (90° - θ) are complementary angles.
(i) sin (90° - θ) = cos θ (iii) tan (90° - θ) = cot θ (v) sec (90° - θ) = csc θ |
(ii) cos (90° - θ) = sin θ (iv) cot (90° - θ) = tan θ (vi) csc (90° - θ) = sec θ |
We know there are six trigonometrical ratios in trigonometry. The above explanation will help us to find the trigonometrical ratios of complementary angles.
Worked-out problems on trigonometrical ratios of complementary angles:
1. Without using trigonometric tables, evaluate \(\frac{tan 65°}{cot 25°}\)
Solution:
\(\frac{tan 65°}{cot 25°}\)
= \(\frac{tan 65°}{cot (90° - 65°)}\)
= \(\frac{tan 65°}{tan 65°}\), [Since cot (90° - θ) = tan θ]
= 1
2. Without using trigonometric tables, evaluate sin 35° sin 55° - cos 35° cos 55°
Solution:
sin 35° sin 55° - cos 35° cos 55°
= sin 35° sin (90° - 35°) - cos 35° cos (90° - 35°),
= sin 35° cos 35° - cos 35° sin 35°,
[Since sin (90° - θ) = cos θ and cos (90° - θ) = sin θ]
= sin 35° cos 35° - sin 35° cos 35°
= 0
3. If sec 5θ = csc (θ - 36°), where 5θ is an acute angle, find the value of θ.
Solution:
sec 5θ = csc (θ - 36°)
⇒ csc (90° - 5θ) = csc (θ - 36°), [Since sec θ = csc (90° - θ)]
⇒ (90° - 5θ) = (θ - 36°)
⇒ -5θ - θ = -36° - 90°
⇒ -6θ = -126°
⇒ θ = 21°, [Dividing both sides by -6]
Therefore, θ = 21°
4. Using trigonometrical ratios of complementary angles prove that tan 1° tan 2° tan 3° ......... tan 89° = 1
Solution:
tan 1° tan 2° tan 3° ...... tan 89°
= tan 1° tan 2° ...... tan 44° tan 45° tan 46° ...... tan 88° tan 89°
= (tan 1° ∙ tan 89°) (tan 2° ∙ tan 88°) ...... (tan 44° ∙ tan 46°) ∙ tan 45°
= {tan 1° ∙ tan (90° - 1°)} ∙ {tan 2° ∙ (tan 90° - 2°)} ...... {tan 44° ∙ tan (90° - 44°)} ∙ tan 45°
= (tan 1° ∙ cot 1°)(tan 2° ∙ cot 2°) ...... (tan 44° ∙ cot 44°) ∙ tan 45°, [Since tan (90° - θ) = cot θ]
= (1)(1) ...... (1) ∙ 1, [since tan θ ∙ cot θ = 1 and tan 45° = 1]
= 1
Therefore, tan 1° tan 2° tan 3° ...... tan 89° = 1
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