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If a1, a2....an are in A.P. with common difference d ¹ 0, then the sum of the series
(sin d) [cosec a1 cosec a2 + cosec a2 cosec a3 +..... ......+ cosec an – 1 cosec an]
- a)sec a1 – sec an
- b)cosec a1 – cosec an
- c)cot a1 – cot an
- d)tan a1 – tan an
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If a1, a2....anare in A.P. with common difference d ¹ 0, then the...
A₁ , A₂ , A₃ , .......An are in AP and common difference is d
so, d = A₂ - A₁ = A₃ - A₂ = A₄ - A₃ =...... =An - An-1 -------(1)
Now, sind[cosecA₁.cosecA₂ + cosecA₂.cosecA₃ + cosecA₃.cosecA₄ ...........+ cosecAn-1.cosecAn ]
⇒ sind[1/sinA₁.sinA₂ + 1/sinA₂.sinA₃ + 1/sinA₃.sinA₄ + .........+ 1/sinAn₋₁.sinAn]
⇒ sin d/sinA₁.sinA₂ + sind/sinA₂.sinA₃ + sind/sinA₃.sinA₄ + .......+sind/sinAn₋₁.sinAn
⇒ sin(A₂ - A₁)/sinA₁.sinA₂ + sin(A₃ - A₂)/sinA₂.sinA₃ + sin(A₄ - A₃)/sinA₃.sinA₄ + .......+ sin(An-1 - An)/sinAn-1.sinAn [ from equation (1)
⇒ (sinA₁.cosA₂ - cosA₁.sinA₂)/sinA₁.sinA₂ + (sinA₂.cosA₃ - cosA₂.sinA₃)/sinA₂.sinA₃ +.........+ (sinAn-1.cosAn - cosAn-1.sinAn)/sinAn-1.sinAn
⇒cotA₂ - cotA₁ + cotA₃ - cotA₂ + cotA₄ - cotA₃ + .........cotAn - cotAn-1
⇒cotAn - cotA₁
so, d = A₂ - A₁ = A₃ - A₂ = A₄ - A₃ =...... =An - An-1 -------(1)
Now, sind[cosecA₁.cosecA₂ + cosecA₂.cosecA₃ + cosecA₃.cosecA₄ ...........+ cosecAn-1.cosecAn ]
⇒ sind[1/sinA₁.sinA₂ + 1/sinA₂.sinA₃ + 1/sinA₃.sinA₄ + .........+ 1/sinAn₋₁.sinAn]
⇒ sin d/sinA₁.sinA₂ + sind/sinA₂.sinA₃ + sind/sinA₃.sinA₄ + .......+sind/sinAn₋₁.sinAn
⇒ sin(A₂ - A₁)/sinA₁.sinA₂ + sin(A₃ - A₂)/sinA₂.sinA₃ + sin(A₄ - A₃)/sinA₃.sinA₄ + .......+ sin(An-1 - An)/sinAn-1.sinAn [ from equation (1)
⇒ (sinA₁.cosA₂ - cosA₁.sinA₂)/sinA₁.sinA₂ + (sinA₂.cosA₃ - cosA₂.sinA₃)/sinA₂.sinA₃ +.........+ (sinAn-1.cosAn - cosAn-1.sinAn)/sinAn-1.sinAn
⇒cotA₂ - cotA₁ + cotA₃ - cotA₂ + cotA₄ - cotA₃ + .........cotAn - cotAn-1
⇒cotAn - cotA₁
Most Upvoted Answer
If a1, a2....anare in A.P. with common difference d ¹ 0, then the...
A₁ , A₂ , A₃ , .......An are in AP and common difference is d
so, d = A₂ - A₁ = A₃ - A₂ = A₄ - A₃ =...... =An - An-1 -------(1)
Now, sind[cosecA₁.cosecA₂ + cosecA₂.cosecA₃ + cosecA₃.cosecA₄ ...........+ cosecAn-1.cosecAn ]
⇒ sind[1/sinA₁.sinA₂ + 1/sinA₂.sinA₃ + 1/sinA₃.sinA₄ + .........+ 1/sinAn₋₁.sinAn]
⇒ sin d/sinA₁.sinA₂ + sind/sinA₂.sinA₃ + sind/sinA₃.sinA₄ + .......+sind/sinAn₋₁.sinAn
⇒ sin(A₂ - A₁)/sinA₁.sinA₂ + sin(A₃ - A₂)/sinA₂.sinA₃ + sin(A₄ - A₃)/sinA₃.sinA₄ + .......+ sin(An-1 - An)/sinAn-1.sinAn [ from equation (1)
⇒ (sinA₁.cosA₂ - cosA₁.sinA₂)/sinA₁.sinA₂ + (sinA₂.cosA₃ - cosA₂.sinA₃)/sinA₂.sinA₃ +.........+ (sinAn-1.cosAn - cosAn-1.sinAn)/sinAn-1.sinAn
⇒cotA₂ - cotA₁ + cotA₃ - cotA₂ + cotA₄ - cotA₃ + .........cotAn - cotAn-1
⇒cotAn - cotA₁
so, d = A₂ - A₁ = A₃ - A₂ = A₄ - A₃ =...... =An - An-1 -------(1)
Now, sind[cosecA₁.cosecA₂ + cosecA₂.cosecA₃ + cosecA₃.cosecA₄ ...........+ cosecAn-1.cosecAn ]
⇒ sind[1/sinA₁.sinA₂ + 1/sinA₂.sinA₃ + 1/sinA₃.sinA₄ + .........+ 1/sinAn₋₁.sinAn]
⇒ sin d/sinA₁.sinA₂ + sind/sinA₂.sinA₃ + sind/sinA₃.sinA₄ + .......+sind/sinAn₋₁.sinAn
⇒ sin(A₂ - A₁)/sinA₁.sinA₂ + sin(A₃ - A₂)/sinA₂.sinA₃ + sin(A₄ - A₃)/sinA₃.sinA₄ + .......+ sin(An-1 - An)/sinAn-1.sinAn [ from equation (1)
⇒ (sinA₁.cosA₂ - cosA₁.sinA₂)/sinA₁.sinA₂ + (sinA₂.cosA₃ - cosA₂.sinA₃)/sinA₂.sinA₃ +.........+ (sinAn-1.cosAn - cosAn-1.sinAn)/sinAn-1.sinAn
⇒cotA₂ - cotA₁ + cotA₃ - cotA₂ + cotA₄ - cotA₃ + .........cotAn - cotAn-1
⇒cotAn - cotA₁
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If a1, a2....anare in A.P. with common difference d ¹ 0, then the...
Understanding the Problem
The problem involves a series of angles that are in Arithmetic Progression (A.P.) with a common difference d. We need to evaluate the expression involving cosecant functions and the sum of their products.
Given
- Angles: a1, a2, ..., an are in A.P.
- Common difference: d > 0
- We need to find the sum: (sin d) [cosec a1 * cosec a2 + cosec a2 * cosec a3 + ... + cosec an-1 * cosec an]
Expression Breakdown
1. Cosecant Function:
- Cosecant is defined as the reciprocal of sine: cosec x = 1/sin x.
2. Using A.P. Properties:
- The angles can be expressed as:
- a1 = A
- a2 = A + d
- a3 = A + 2d
- ...
- an = A + (n-1)d
3. Summation:
- The expression simplifies to a sum of products of cosecants.
Key Insight
Using trigonometric identities, particularly the product-to-sum identities, we can relate these products to cotangent functions. The sum of products of cosecants can be simplified to yield cotangent differences.
Final Result
After simplification, we find that the entire expression results in:
- cot a1 - cot an
This matches with option 'C'.
Conclusion
Thus, the correct answer to the problem is option 'C': cot a1 - cot an. This showcases the power of utilizing properties of A.P. and trigonometric identities for simplifying complex expressions.
The problem involves a series of angles that are in Arithmetic Progression (A.P.) with a common difference d. We need to evaluate the expression involving cosecant functions and the sum of their products.
Given
- Angles: a1, a2, ..., an are in A.P.
- Common difference: d > 0
- We need to find the sum: (sin d) [cosec a1 * cosec a2 + cosec a2 * cosec a3 + ... + cosec an-1 * cosec an]
Expression Breakdown
1. Cosecant Function:
- Cosecant is defined as the reciprocal of sine: cosec x = 1/sin x.
2. Using A.P. Properties:
- The angles can be expressed as:
- a1 = A
- a2 = A + d
- a3 = A + 2d
- ...
- an = A + (n-1)d
3. Summation:
- The expression simplifies to a sum of products of cosecants.
Key Insight
Using trigonometric identities, particularly the product-to-sum identities, we can relate these products to cotangent functions. The sum of products of cosecants can be simplified to yield cotangent differences.
Final Result
After simplification, we find that the entire expression results in:
- cot a1 - cot an
This matches with option 'C'.
Conclusion
Thus, the correct answer to the problem is option 'C': cot a1 - cot an. This showcases the power of utilizing properties of A.P. and trigonometric identities for simplifying complex expressions.
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If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer?
Question Description
If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer?.
If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer?.
Solutions for If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice If a1, a2....anare in A.P. with common difference d ¹ 0, then the sum of the series(sin d) [cosec a1cosec a2+ cosec a2cosec a3+..... ......+ cosec an –1cosec an]a)sec a1–sec anb)cosec a1–cosec anc)cot a1–cot and)tan a1–tan anCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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