# Trigonometric Ratios of Complementary Angles

Complementary angles and their trigonometric ratios:

We know from geometry if the sum of two angles is 90°, then one angle is called the complement of the other.

Two angles A and B are complementary if A + B = 90°. So, B = 90° - A.

For example, as 30° + 60° = 90°, 60° is called the complement of 30° and conversely, 30° is called the complement of 60°.

Thus 27° is the complement of 60°; 43.5° is the complement of 46.5° etc.

Thus in general, (90° - θ) and θ are complementary angles. Trigonometric ratios of (90° - θ) are convertible to trigonometric ratios of θ.

Trigonometric Ratios of 90° - θ in Terms of Trigonometric ratios of θ

Let us see how we can find the trigonometrical ratios of 90° - θ, if we know those of θ°.

Let PQR be a right-angled triangle in which ∠Q is the right angle.

Let ∠PRQ = θ. Then, ∠QPR = 180° - (90° + θ) = 90° - θ.

1. sin (90° - θ) = cos θ

Here, sin (90° - θ) = $$\frac{QR}{PR}$$ and cos θ = $$\frac{QR}{PR}$$

Therefore, sin (90° - θ) = cos θ.

2. cos (90° - θ) = sin θ

Here, cos (90° - θ) = $$\frac{PQ}{PR}$$ and sin θ = $$\frac{PQ}{PR}$$

Therefore, cos (90° - θ) = sin θ.

3. tan (90° - θ) = cot θ

Here, tan (90° - θ) = $$\frac{QR}{PQ}$$ and cot θ = $$\frac{QR}{PQ}$$

Therefore, tan (90° - θ) = cot θ.

4. csc (90° - θ) = sec θ

Here, csc (90° - θ) = $$\frac{PR}{QR}$$ and sec θ = $$\frac{PR}{QR}$$

Therefore, csc (90° - θ) = sec θ

5. sec (90° - θ) = csc θ

Here, sec (90° - θ) = $$\frac{PR}{PQ}$$ and csc θ = $$\frac{PR}{PQ}$$

Therefore, sec (90° - θ) = csc θ.

6. cot (90° - θ) = tan θ

Here, cot (90° - θ) = $$\frac{PQ}{QR}$$ and tan θ = $$\frac{PQ}{QR}$$

Therefore, cot (90° - θ) = tan θ.

Thus, we have the following conversions of trigonometric ratios of (90° - θ) in terms of trigonometric ratios of θ.

 sin (90° - θ) = cos θcos (90° - θ) = sin θ tan (90° - θ) = cot θcot (90° - θ) = tan θ sec (90° - θ) = csc θcsc (90° - θ) = sec θ

For example, cos 37° can be expressed as sine of the complementary angle of 37° because

cos 37° = cos (90° - 53°) = sin 53°.

Note: The measure of an angle can be expressed in degrees (°) as well as in radians. The measure of an angle is π radians (where π is 3.14, approximately) if its measure in degrees is 180°. Thus, 180° = π radians. This is also written as 180° = π.

Therefore, 1° = $$\frac{π}{180}$$

30° = $$\frac{π}{6}$$

45° = $$\frac{π}{4}$$

60° = $$\frac{π}{3}$$

90° = $$\frac{π}{2}$$, etc.

Therefore, we can write sin (90° - β) = sin ($$\frac{π}{2}$$ – β) = cos β

cos (90° - β) = cos ($$\frac{π}{2}$$ – β) = sin β

tan (90° - β) = tan ($$\frac{π}{2}$$ – β) = cot β

csc (90° - β) = csc ($$\frac{π}{2}$$ – β) = sec β

sec (90° - β) = sec ($$\frac{π}{2}$$ – β) = csc β

cot (90° - β) = cot ($$\frac{π}{2}$$ – β) = tan β.

The values of trigonometrical ratios of 30° and 60°, which are complementary angles are compared below. This will help us to have a clear understanding of the relations shown before.

sin 30° = cos 60° = $$\frac{1}{2}$$

cos 30° = sin 60° = $$\frac{\sqrt{3}}{2}$$

tan 30° = cot 60° = $$\frac{\sqrt{3}}{3}$$

csc 30° = sec 60° = 2

sec 30° = csc 60° = $$\frac{2\sqrt{3}}{3}$$

cot 30° = tan 60° = $$\sqrt{3}$$

Similarly, from the complementary angles formulae we get

sin 45° = cos 45° = $$\frac{\sqrt{2}}{2}$$

tan 45° = cot 45° = 1

csc 45 = sec 45° = $$\sqrt{2}$$

tan 45° = cot 45° = 1

Again,

sin 90° = cos 0° = 1

cos 90° = sin 0° = 0

Problems on Trigonometric Ratios of Complementary Angles

Problems on evaluation using trigonometric ratios of complementary angles

1. Evaluate without using trigonometric table: $$\frac{sin 25°}{2 ∙ cos 65°}$$

Solution:

$$\frac{sin 25°}{2 ∙ cos 65°}$$

= $$\frac{sin 25°}{2 ∙ cos (90° - 25°)}$$

= $$\frac{sin 25°}{2 ∙ sin 25°}$$; [since, cos (90° - θ) = sin θ]

= $$\frac{1}{2}$$.

2. Evaluate without using trigonometric table: tan 38° ∙ tan 52°

Solution:

tan 38° ∙ tan 52°

= tan 38° ∙ tan (90° - 38°)

= tan 38° ∙ cot 38°; [Since, tan (90° - θ) = cot θ]

= tan 38° ∙ $$\frac{1}{tan 38°}$$

= 1.

3. Evaluate without using trigonometric table: $$\frac{sin 67°}{cos 23°}$$ - $$\frac{sec 12°}{csc 78°}$$

Solution:

$$\frac{sin 67°}{cos 23°}$$ - $$\frac{sec 12°}{csc 78°}$$

= $$\frac{sin 67°}{cos (90° - 67°)}$$ - $$\frac{sec 12°}{csc (90° - 12°)}$$

= $$\frac{sin 67°}{cos (90° - 67°)}$$ - $$\frac{sec 12°}{csc (90° - 12°)}$$

= $$\frac{sin 67°}{sin 67°}$$ - $$\frac{sec 12°}{sec 12°}$$

[Since, cos (90° - θ) = sin θ and csc (90° - θ) = sec θ]

= 1 - 1

= 0.

4. If cos 39° = $$\frac{x}{\sqrt{x^{2} + y^{2}}}$$, what is the value of tan 51°?

Solution:

Given that cos 39° = $$\frac{x}{\sqrt{x^{2} + y^{2}}}$$

Therefore, sin2 39° = 1 -  $$\frac{x^{2}}{x^{2} + y^{2}}$$

= $$\frac{x^{2} + y^{2} - x^{2}}{x^{2} + y^{2}}$$

= $$\frac{y^{2}}{x^{2} + y^{2}}$$

Therefore, sin 39° = $$\frac{y}{\sqrt{x^{2} + y^{2}}}$$, (negative value is not acceptable)

Now, tan 51° = tan (90° - 39°)

=  cot 39°

= $$\frac{cos 39°}{sin 39°}$$

= cos 39° ÷ sin 39°

= $$\frac{x}{\sqrt{x^{2} + y^{2}}}$$ ÷ $$\frac{y}{\sqrt{x^{2} + y^{2}}}$$

= $$\frac{x}{y}$$.

5. If cos 37° = x then find the value of tan 53°.

Solution:

tan 53°

= tan (90° - 37°)

= cot 37°; [Since, tan (90° - θ) = cot θ]

= $$\frac{cos 37°}{sin 37°}$$

= $$\frac{x}{sin 37°}$$ ................ (i)

Now, sin2 37° = 1 - cos2 37°; [since, 1 - cos2 θ = sin2 θ]

Therefore, sin 37° = $$\sqrt{1 - cos^{2} 37°}$$

= $$\sqrt{1 - x^{2}}$$

Therefore, from (i), tan 53° = $$\frac{x}{\sqrt{1 - x^{2}}}$$.

6. If sec  ϕ = csc  β and 0° < (ϕ, β) < 90°, find the value of sin (ϕ + β).

Solution:

sec  ϕ = csc  β

$$\frac{1}{cos ϕ}$$ = $$\frac{1}{sin β}$$

⟹ cos ϕ = sin β

⟹ cos ϕ = cos (90° - β)

⟹ ϕ = 90° - β

⟹ ϕ + β = 90°

Therefore, sin (ϕ + β) = sin 90° = 1.

7. Find the value of sin2 15° + sin2 25° + sin2 33° + sin2 57° + sin2 65° + sin2 75°.

Solution:

sin2 (90° - 75°) + sin2 (90° - 65°) + sin2 (90° - 57°) + sin2 57° + sin2 65° + sin2 75°.

= cos2 75° + cos2 65° + cos2 57° + sin2 57° + sin2 65° + sin2 75°.

= (sin2 57° + cos2 75°) + (sin2 65° + cos2 65°) + (sin2 57° + cos2 57°)

= 1 + 1 + 1; [Since, sin2 θ + cos2 θ = 1]

= 3.

8. If tan 49° ∙ cot (90° - θ) = 1, find θ.

Solution:

tan 49° ∙ cot (90° - θ) = 1

⟹ tan 49° ∙ tan θ = 1; [Since, cot (90° - θ) = tan θ]

⟹ tan θ = $$\frac{1}{tan 49°}$$

⟹ tan θ = cot 49°

⟹ tan θ = cot (90° - 41°)

⟹ tan θ = tan 41°

⟹ θ = 41°

Therefore, θ = tan 41°.

Problems on establishing equality using trigonometric ratios of complementary angles

9. Prove that sin 33° cos 77° = cos 57° sin 13°

Solution:

LHS = sin 33° cos 77°

= sin (90° - 57°) cos (90° - 13°)

= cos 57° sin 13°

= RHS. (Proved).

10. Prove that tan 11° + cot 63° = tan 27° + cot 79°

Solution:

LHS = tan 11° + cot 63°

= tan (90° - 79°) + cot (90° - 27°)

= cot 79° + tan 27°

= tan 27° + cot 79°

= RHS. (Proved).

Problems on establishing identities and simplification using trigonometric ratios of complementary angles

11. If P and Q are two complementary angles, show that

(sin P + sin Q)2 = 1 + 2 sin P cos P

Solution:

Since P are Q are complementary angles,

Therefore, sin Q = sin (90° - P) = cos P

Therefore, (sin P + sin Q)2 = (sin P + cos P)2

= sin2 P + cos2 P + 2 sin P cos P

= (sin2 P + cos2 P) + 2 sin P cos P

= 1 + 2 sin P cos P

12. Simplify: $$\frac{sin (\frac{π}{2} - θ) ∙ cot (\frac{π}{2} - θ)}{sin θ}$$

Solution:

$$\frac{sin (\frac{π}{2} - θ) ∙ cot (\frac{π}{2} - θ)}{sin θ}$$

= $$\frac{cos θ ∙ tan θ}{sin θ}$$, [Since sin ($$\frac{π}{2}$$ - θ) = sin (90° - θ) = cos θ and cot ($$\frac{π}{2}$$ - θ) = cot (90° - θ) = tan θ]

= $$\frac{cos θ ∙ \frac{sin θ}{cos θ}}{sin θ}$$

=  $$\frac{sin θ}{sin θ}$$

= 1.

13. Prove that, sin2 7° + sin2 83°

Solution:

sin 83° = sin (90° - 7°)

= cos 7°; [since, sin (90° - θ) = cos θ]

LHS = sin2 7° + sin2 83°

= sin2 7° + cos2 7°, [Since, sin 83° = cos 7°]

= 1 = RHS (Proved).

14. In a ∆PQR, prove that sin $$\frac{P + Q}{2}$$ = cos $$\frac{R}{2}$$.

Solution:

We know that sum of the three angles of a triangle is 180°.

i,e., P + Q + R = 180°

⟹  P + Q = 180° - R

Now,

LHS = sin $$\frac{P + Q}{2}$$

= sin $$\frac{180° - R}{2}$$

= sin (90° - $$\frac{R}{2}$$)

= cos $$\frac{R}{2}$$ = RHS (Proved).

15. Prove that tan 15° + tan 75° = $$\frac{sec^{2} 15°}{\sqrt{sec^{2} 15° - 1}}$$.

Solution:

LHS = tan 15° + tan (90° - 15°)

= tan 15° + cot 15°

= tan 15° + $$\frac{1}{tan 15°}$$

= $$\frac{tan^{2} 15° + 1}{tan 15°}$$

= $$\frac{sec^{2} 15°}{\sqrt{sec^{2} 15° - 1}}$$ = RHS (Proved).

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