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In a triangle ABC,cosA- cosB=
- a)2cos(C/2) sin(A+B/2)
- b)-2sin(C/2) sin(A+B/2)
- c)– 2sin(C/2) sin(A-B/2)
- d)-2cos(C/2) sin(A-B)/2
Correct answer is option 'D'. Can you explain this answer?
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In a triangle ABC,cosA- cosB=a)2cos(C/2) sin(A+B/2)b)-2sin(C/2) sin(A+...
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In a triangle ABC,cosA- cosB=a)2cos(C/2) sin(A+B/2)b)-2sin(C/2) sin(A+...
Explanation:
Given:
Triangle ABC with cosA - cosB
To prove:
-2cos(C/2) sin(A-B)/2
Steps:
1. Use the trigonometric identity: cosC = cos(A-B) = cosA cosB + sinA sinB
2. Substitute the given values of cosA and cosB into the identity
3. Simplify the expression using trigonometric identities for sine and cosine functions
4. Factor out -2sin(C/2) from the expression to get the desired result
Therefore, the correct answer is option 'D': -2cos(C/2) sin(A-B)/2.
Given:
Triangle ABC with cosA - cosB
To prove:
-2cos(C/2) sin(A-B)/2
Steps:
1. Use the trigonometric identity: cosC = cos(A-B) = cosA cosB + sinA sinB
2. Substitute the given values of cosA and cosB into the identity
3. Simplify the expression using trigonometric identities for sine and cosine functions
4. Factor out -2sin(C/2) from the expression to get the desired result
Therefore, the correct answer is option 'D': -2cos(C/2) sin(A-B)/2.
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In a triangle ABC,cosA- cosB=a)2cos(C/2) sin(A+B/2)b)-2sin(C/2) sin(A+...
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In a triangle ABC,cosA- cosB=a)2cos(C/2) sin(A+B/2)b)-2sin(C/2) sin(A+B/2)c)– 2sin(C/2) sin(A-B/2)d)-2cos(C/2) sin(A-B)/2Correct answer is option 'D'. Can you explain this answer?
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