Trigonometrical Ratios of Complementary Angles

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

Trigonometrical Ratios of Complementary Angles

\(\overrightarrow{OX}\) 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

11 and 12 Grade Math

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