Evaluate each of the following :
Problem 1 :
sin(180° - θ)
Problem 2 :
cos(180° - θ)
Problem 3 :
tan(180° - θ)
Problem 4 :
csc(180° - θ)
Problem 5 :
sec(180° - θ)
Problem 6 :
cot(180° - θ)
Problem 7 :
sin(180° + θ)
Problem 8 :
cos(180° + θ)
Problem 9 :
tan(180° + θ)
Problem 10 :
csc(180° + θ)
Problem 11 :
sec(180° + θ)
Problem 12 :
cot(180° + θ)
1. Answer :
sin(180° - θ)
To evaluate sin(180° - θ), we have to consider the following important points.
(i) (180° - θ) will fall in the II nd quadrant.
(ii) When we have 180°, "sin" will not be changed as "cos".
(iii) In the II nd quadrant, the sign of "sin" is positive.
Considering the above points, we have
sin(180° - θ) = sinθ
2. Answer :
cos(180° - θ)
To evaluate cos(180° - θ), we have to consider the following important points.
(i) (180° - θ) will fall in the II nd quadrant.
(ii) When we have 180°, "cos" will not be changed as "sin".
(iii) In the II nd quadrant, the sign of "cos" is negative.
Considering the above points, we have
cos(180° - θ) = -cosθ
3. Answer :
tan(180° - θ)
To evaluate tan(180° - θ), we have to consider the following important points.
(i) (180° - θ) will fall in the II nd quadrant.
(ii) When we have 180°, "tan" will not be changed as "cot".
(iii) In the II nd quadrant, the sign of "tan" is negative.
Considering the above points, we have
tan(180° - θ) = -tanθ
4. Answer :
csc(180° - θ)
To evaluate csc(180° - θ), we have to consider the following important points.
(i) (180° - θ) will fall in the II nd quadrant.
(ii) When we have 180°, "csc" will not be changed as "sec".
(iii) In the II nd quadrant, the sign of "csc" is positive.
Considering the above points, we have
csc(180° - θ) = cscθ
5. Answer :
sec(180° - θ)
To evaluate sec(180° - θ), we have to consider the following important points.
(i) (180° - θ) will fall in the II nd quadrant.
(ii) When we have 180°, "sec" will not be changed as "csc".
(iii) In the II nd quadrant, the sign of "sec" is negative.
Considering the above points, we have
sec(180° - θ) = -secθ
6. Answer :
cot(180° - θ)
To evaluate cot(180° - θ), we have to consider the following important points.
(i) (180° - θ) will fall in the II nd quadrant.
(ii) When we have 180°, "cot" will not be changed as "tan".
(iii) In the II nd quadrant, the sign of "cot" is negative.
Considering the above points, we have
cot(180° - θ) = -cotθ
7. Answer :
sin(180° + θ)
To evaluate sin(180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "sin" will not be changed as "cos".
(iii) In the III rd quadrant, the sign of "sin" is negative.
Considering the above points, we have
sin(180° + θ) = -sinθ
8. Answer :
cos(180° + θ)
To evaluate cos(180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "cos" will not be changed as "sin".
(iii) In the III rd quadrant, the sign of "cos" is negative.
Considering the above points, we have
cos(180° + θ) = -cosθ
9. Answer :
tan(180° + θ)
To evaluate tan(180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "tan" will not be changed as "cot".
(iii) In the III rd quadrant, the sign of "tan" is positive.
Considering the above points, we have
tan(180° + θ) = tanθ
10. Answer :
csc(180° + θ)
To evaluate csc(180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "csc" will not be changed as "sec".
(iii) In the III rd quadrant, the sign of "csc" is negative.
Considering the above points, we have
csc(180° + θ) = -cscθ
11. Answer :
sec(180° + θ)
To evaluate sec(180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "sec" will not be changed as "csc".
(iii) In the III rd quadrant, the sign of "sec" is negative.
Considering the above points, we have
sec(180° + θ) = -secθ
12. Answer :
cot(180° + θ)
To evaluate cot(180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "cot" will not be changed as "tan".
(iii) In the III rd quadrant, the sign of "cot" is positive.
Considering the above points, we have
cot(180° + θ) = cotθ
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