TRIGONOMETRIC RATIOS OF 90 DEGREE PLUS THETA 

About "Trigonometric Ratios of 90 Degree Plus Theta"

Trigonometric ratios of 90 degree plus theta is one of the branches of ASTC formula in trigonometry. 

Trigonometric-ratios of 90 degree plus theta are given below.

sin (90° + θ)  =  cos θ

cos (90° + θ)  =  - sin θ

tan (90° + θ)  =  - cot θ

csc (90° + θ)  =  sec θ

sec (90° + θ)  =  - csc θ

cot (90° + θ)  =  - tan θ

Let us see, how the trigonometric ratios of 90 degree plus theta are determined. 

To know that, first we have to understand ASTC formula. 

The ASTC formula can be remembered easily using the following phrases.

"All Sliver Tea Cups" 

or

"All Students Take Calculus"

ASTC formla has been explained clearly in the figure given below.

More clearly 

From the above picture, it is very clear that 

(90° + θ) falls in the second quadrant

In the second quadrant (90° + θ), sin and csc are positive and other trigonometric ratios are negative.

Important conversions

When we have the angles 90° and 270° in the trigonometric ratios in the form of

(90° + θ)

(90° - θ)

(270° + θ)

(270° - θ)

We have to do the following conversions, 

sin θ <------> cos θ

tan θ <------> cot θ

csc θ <------> sec θ

For example,

sin (270° + θ)  =  - cos θ

cos (90° - θ)  =  sin θ

For the angles 0° or 360° and  180°, we should not make the above conversions. 

Evaluation of Trigonometric Ratios using ASTC formula

Problem 1 :

Evaluate : sin (90° + θ)

Solution :

To evaluate sin (90° + θ), we have to consider the following important points. 

(i)  (90° + θ) will fall in the II nd quadrant. 

(ii)  When we have 90°, "sin" will become "cos".

(iii)  In the II nd quadrant, the sign of "sin" is positive. 

Considering the above points, we have 

sin (90° + θ)  =  cos θ

Problem 2 :

Evaluate : cos (90° + θ)

Solution :

To evaluate cos (90° + θ), we have to consider the following important points. 

(i)  (90° + θ) will fall in the II nd quadrant. 

(ii)  When we have 90°, "cos" will become "sin".

(iii)  In the II nd quadrant, the sign of "cos" is negative. 

Considering the above points, we have 

cos (90° + θ)  =  - sin θ

Problem 3 :

Evaluate : tan (90° + θ)

Solution :

To evaluate tan (90° + θ), we have to consider the following important points. 

(i)  (90° + θ) will fall in the II nd quadrant. 

(ii)  When we have 90°, "tan" will become "cot".

(iii)  In the II nd quadrant, the sign of "tan" is negative. 

Considering the above points, we have 

tan (90° + θ)  =  - cot θ

Problem 4 :

Evaluate : csc (90° + θ)

Solution :

To evaluate csc (90° + θ), we have to consider the following important points. 

(i)  (90° + θ) will fall in the II nd quadrant. 

(ii)  When we have 90°, "csc" will become "sec".

(iii)  In the II nd quadrant, the sign of "csc" is positive. 

Considering the above points, we have 

csc (90° + θ)  =  sec θ

Problem 5 :

Evaluate : sec (90° + θ)

Solution :

To evaluate sec (90° + θ), we have to consider the following important points. 

(i)  (90° + θ) will fall in the II nd quadrant. 

(ii)  When we have 90°, "sec" will become "csc".

(iii)  In the II nd quadrant, the sign of "sec" is negative. 

Considering the above points, we have 

sec (90° + θ)  =  - csc θ

Problem 6 :

Evaluate : cot (90° + θ)

Solution :

To evaluate cot (90° + θ), we have to consider the following important points. 

(i)  (90° + θ) will fall in the II nd quadrant. 

(ii)  When we have 90°, "cot" will become "tan"

(iii)  In the II nd quadrant, the sign of "cot" is negative. 

Considering the above points, we have 

cot (90° + θ)  =  - tan θ

Summary (90 degree plus theta)

sin (90° + θ)  =  cos θ

cos (90° + θ)  =  - sin θ

tan (90° + θ)  =  - cot θ

csc (90° + θ)  =  sec θ

sec (90° + θ)  =  - csc θ

cot (90° + θ)  =  - tan θ

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