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.

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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