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 again the same rotating line rotates in the same direction and makes an angle ∠AOB =90°.

Therefore we see that, ∠XOB = 90° + θ.

Take a point C on OA and draw CD perpendicular to OX or OX’.

Again, take a point E on OB such that OE = OC and draw EF perpendicular to OX or OX’. From the right-angled ∆ OCD and ∆ OEF we get,

∠COD = ∠OEF [since OB ⊥ OA]

and OC = OE.

Therefore, ∆ OCD ≅ ∆ OEF (congruent).

Therefore according to the definition of trigonometric sign, OF = - DC, FE = OD and OE = OC

We observe that in diagram 1 and 4 OF and DC are opposite signs and FE, OD are either both positive. Again we observe that in diagram 2 and 3 OF and DC are opposite signs and FE, OD 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 ∆ OCD ≅ ∆ OEF]

**sin (90° + θ) = cos θ**

cos (90° + θ) = \(\frac{OF}{OE}\)

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

**cos (90° + θ) = - sin θ.**

tan (90° + θ) = \(\frac{FE}{OF}\)

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

**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 sin 135°.

**Solution:**

sin 135° = sin (90 + 45)°

= cos 45°; since we know, sin (90° + θ) = cos θ

= \(\frac{1}{√2}\)

**2.** Find the value of tan 150°.

**Solution:**

tan 150° = tan (90 + 60)°

= - cot 60°; since we know, tan (90° + θ) = - cot θ

= \(\frac{1}{√3}\)

**●** **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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