Trigonometrical Ratios of (90° - θ)

What is the relation among all the trigonometrical ratios of (90° - θ)?

In trigonometrical ratios of angles (90° - θ) we will find the relation between all six trigonometrical ratios.

Let a rotating line OA rotates about O in the anti-clockwise direction, from initial position to ending position makes an angle ∠XOA = θ. Now a point C is taken on OA and draw CD perpendicular to OX or OX'.

Again another rotating line OB rotates about O in the anti-clockwise direction, from initial position to ending position (OX) makes an angle ∠XOY = 90°; this rotating line now rotates in the clockwise direction, starting from the position (OY) makes an angle ∠YOB = θ.

Now, we can observe that ∠XOB = 90° - θ.

Again a point E is taken on OB such that OC = OE and draw EF perpendicular to 

OX or OX'.

Since, ∠YOB = ∠XOA

Therefore, ∠OEF = ∠COD.

Now, from the right-angled ∆EOF and right-angled ∆COD we get, ∠OEF = ∠COD and OE = OC.

Hence, ∆EOF ≅ ∆COD (congruent).

Therefore, FE = OD, OF = DC and OE = OC.

In this diagram FE and OD both are positive. Similarly, OF and DC are both positive.

In this diagram FE and OD both are negative. Similarly, OF and DC are both negative.

In this diagram FE and OD both are negative. Similarly, OF and DC are both negative.

In this diagram FE and OD both are positive. Similarly, OF and DC are both negative.

According to the definition of trigonometric ratio we get,

sin (90° - θ) = \(\frac{FE}{OE}\)

sin (90° - θ) = \(\frac{OD}{OC}\), [FE = OD and OE = OC, since ∆EOF ≅ ∆COD]

sin (90° - θ) = cos θ

cos (90° - θ) = \(\frac{OF}{OE}\)

cos (90° - θ) = \(\frac{DC}{OC}\), [OF = DC and OE = OC, since ∆EOF ≅ ∆COD]

cos (90° - θ) = sin θ

tan (90° - θ) = \(\frac{FE}{OF}\)

tan (90° - θ) = \(\frac{OD}{DC}\), [FE = OD and OF = DC, since ∆EOF ≅ ∆COD]

tan (90° - θ) = cot θ

Similarly, csc (90° - θ) = \(\frac{1}{sin (90°  -  \Theta)}\) 

csc (90° - θ) = \(\frac{1}{cos \Theta}\)

csc (90° - θ) = sec θ

sec ( 90° - θ) = \(\frac{1}{cos (90°  -  \Theta)}\) 

sec (90° - θ) = \(\frac{1}{sin \Theta}\)

sec (90° - θ) = csc θ

and cot (90° - θ) = \(\frac{1}{tan (90°  -  \Theta)}\) 

cot (90° - θ) = \(\frac{1}{cot \Theta}\)

cot (90° - θ) = tan θ

Solved examples:

1. Find the value of cos 30°.

Solution:

cos 30° = sin (90 - 60)°

            = sin 60°; since we know, cos (90° - θ) = sin θ

              = \(\frac{√3}{2}\)

2. Find the value of csc 90°.

Solution:

csc 90° = csc (90 - 0)°

            = sec 0°; since we know, csc (90° - θ) = sec θ

              = 1

● Trigonometric Functions

• Basic Trigonometric Ratios and Their Names
• Restrictions of Trigonometrical Ratios
• Reciprocal Relations of Trigonometric Ratios
• Quotient Relations of Trigonometric Ratios
  
• Limit of Trigonometric Ratios
  
• Trigonometrical Identity
• Problems on Trigonometric Identities
• Elimination of Trigonometric Ratios
• Eliminate Theta between the equations
• Problems on Eliminate Theta
• Trig Ratio Problems
• Proving Trigonometric Ratios
• Trig Ratios Proving Problems
• Verify Trigonometric Identities
• Trigonometrical Ratios of 0°
• Trigonometrical Ratios of 30°
• Trigonometrical Ratios of 45°
• Trigonometrical Ratios of 60°
• Trigonometrical Ratios of 90°
• Trigonometrical Ratios Table
• Problems on Trigonometric Ratio of Standard Angle
  
• Trigonometrical Ratios of Complementary Angles
• Rules of Trigonometric Signs
• Signs of Trigonometrical Ratios
  
• All Sin Tan Cos Rule
• Trigonometrical Ratios of (- θ)
• Trigonometrical Ratios of (90° + θ)
• Trigonometrical Ratios of (90° - θ)
• Trigonometrical Ratios of (180° + θ)
• Trigonometrical Ratios of (180° - θ)
• Trigonometrical Ratios of (270° + θ)
• Trigonometrical Ratios of (270° - θ)
• Trigonometrical Ratios of (360° + θ)
• Trigonometrical Ratios of (360° - θ)
• Trigonometrical Ratios of any Angle
• Trigonometrical Ratios of some Particular Angles
• Trigonometric Ratios of an Angle
• Trigonometric Functions of any Angles
• Problems on Trigonometric Ratios of an Angle
• Problems on Signs of Trigonometrical Ratios

11 and 12 Grade Math

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