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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