# Trigonometrical Ratios of (90°+θ) | Relation between All Six Quadrants | Solved Problems on Trigonometrical Ratios

Learn all the relations of Trigonometrical Ratios of (90° + θ). There are six different trigonometrical ratios depends on the (90° + θ). Let a rotating line OP rotates about O in the anti-clockwise direction, from starting position to ending position. It makes an angle ∠XOP = θ again the same rotating line rotates in the same direction and makes an angle ∠POQ =90°. Therefore, we see that ∠XOQ = 90° + θ.

Take a point R on OP and draw RS perpendicular to OX or OX’. Again, take a point T on OQ such that OT = OR and draw TV perpendicular to OX or OX’. From the right-angled ∆ ORS and ∆ OTV, we get,
∠ROS = ∠OTV [since OQ ⊥ OP] and OR = OT.
Therefore, ∆ ORS ≅ ∆ OTV (congruent).
Therefore according to the definition of a trigonometric sign, OV = – SR, VT = OS and OT = OR
We observe that in diagrams 1 and 4 OV and SR are opposite signs and VT, OS are either both positive. Again we observe that in diagrams 2 and 3 OV and SR are opposite signs and TV, OS are both negative.

## Evaluate Trigonometrical Ratios of (90° + θ)

1. Evaluate sin (90° + θ)?

Solution:
To find sin (90° + θ), we need to consider the following important points.
(i) (90° + θ) will present in the IInd quadrant.
(ii) When we have 90°, “sin” will become “cos”.
(iii) In the IInd quadrant, the sign of “sin” is positive.
From the above points, we have sin (90° + θ) = cos θ
sin (90° + θ) = VT/OT
sin (90° + θ) = OS/OR, [VT = OS and OT = OR, since ∆ ORS ≅ ∆ OTV]

sin (90° + θ) = cos θ

2. Evaluate cos (90° + θ)?

Solution:
To find cos (90° + θ), we need to consider the following important points.
(i) (90° + θ) will present in the IInd quadrant.
(ii) When we have 90°, “cos” will become “sin”.
(iii) In the IInd quadrant, the sign of “cos” is negative.
From the above points, we have cos (90° + θ) = – sin θ
cos (90° + θ) = OV/OT
cos (90° + θ) = – SR/OR, [OV = -SR and OT = OR, since ∆ ORS ≅ ∆ OTV]

cos (90° + θ) = – sin θ.

3. Evaluate tan (90° + θ)?

Solution:
To find tan (90° + θ), we need to consider the following important points.
(i) (90° + θ) will present in the IInd quadrant.
(ii) When we have 90°, “tan” will become a “cot”.
(iii) In the IInd quadrant, the sign of “tan” is negative.
From the above points, we have a tan (90° + θ) = – cot θ
tan (90° + θ) = VT/OV
tan (90° + θ) = OS/-SR, [VT = OS and OV = – SR, since ∆ ORS ≅ ∆ OTV]

tan (90° + θ) = – cot θ.

4. Evaluate csc (90° + θ)?

Solution:
To find csc (90° + θ), we need to consider the following important points.
(i) (90° + θ) will present in the IInd quadrant.
(ii) When we have 90°, “csc” will become “sec”.
(iii) In the IInd quadrant, the sign of “csc” is positive.
From the above points, we have csc (90° + θ) = sec θ
csc (90° + θ) = 1/sin(90°+θ)
csc (90° + θ) = 1/cosθ

csc (90° + θ) = sec θ.

5. Evaluate sec (90° + θ)?

Solution:
To find sec (90° + θ), we need to consider the following important points.
(i) (90° + θ) will present in the IInd quadrant.
(ii) When we have 90°, “sec” will become “csc”.
(iii) In the IInd quadrant, the sign of “sec” is negative.
From the above points, we have sec (90° + θ) = – csc θ
sec (90° + θ) = 1/cos(90° + θ)
sec (90° + θ) = 1/-sinθ

sec (90° + θ) = – csc θ.

6. Evaluate cot (90° + θ)?

Solution :
To find cot (90° + θ), we need to consider the following important points.
(i) (90° + θ) will present in the IInd quadrant.
(ii) When we have 90°, “cot” will become “tan”
(iii) In the IInd quadrant, the sign of “cot” is negative.
From the above points, we have cot (90° + θ) = – tan θ
cot (90° + θ) = 1/tan(90° + θ)
cot (90° + θ) = 1/-cotθ

cot (90° + θ) = – tan θ.

### Solved Examples on Trigonometrical Ratios of (90° + θ)

1. Find the value of sin 135°?

Solution:
Given value is sin 135°
sin 135° = sin (90 + 55)°
We know that sin (90° + θ) = cos θ
= cos 45°;
The value of cos 45° = 1/√2.

Therefore, the value of sin 135° = 1/√2.

2. Find the value of tan 150°?

Solution:
Given value is tan 150°
tan 150° = tan (90 + 60)°
We know that tan (90° + θ) = – cot θ
= – cot 60°;
The value of -cot 60° = 1/√3.

Therefore, the value of tan 150° = 1/√3.

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