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If x = cosecθ – sinθ and y = secθ – cosθ, then the value of x2y2 (x2 + y2 + 3) is (SSC CGL 1st Sit. 2012)
- a)0
- b)1
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If x = cosecθ – sinθ and y = secθ – cos&...
Given:
x = cosecθ – sinθ
y = secθ – cosθ
To find:
x^2y^2(x^2 + y^2 + 3)
Solution:
Step 1: Finding x^2 and y^2
x^2 = (cosecθ – sinθ)^2
= cosec^2θ + sin^2θ - 2cosecθsinθ
= 1 + 1 - 2cosecθsinθ
= 2 - 2cosecθsinθ
y^2 = (secθ – cosθ)^2
= sec^2θ + cos^2θ - 2secθcosθ
= 1 + 1 - 2secθcosθ
= 2 - 2secθcosθ
Step 2: Finding x^2y^2(x^2 + y^2 + 3)
x^2y^2(x^2 + y^2 + 3) = (2 - 2cosecθsinθ)(2 - 2secθcosθ)(2 - 2cosecθsinθ + 2 - 2secθcosθ + 3)
= 4(1 - cosecθsinθ)(1 - secθcosθ)(7 - 2cosecθsinθ - 2secθcosθ)
Now, 1 - cosecθsinθ = 1 - 1 = 0
1 - secθcosθ = 1 - 1 = 0
Therefore, the expression becomes:
4(0)(0)(7 - 0 - 0)
= 4(0)(7)
= 0
Therefore, x^2y^2(x^2 + y^2 + 3) = 0
Hence, the correct answer is option B - 0.
x = cosecθ – sinθ
y = secθ – cosθ
To find:
x^2y^2(x^2 + y^2 + 3)
Solution:
Step 1: Finding x^2 and y^2
x^2 = (cosecθ – sinθ)^2
= cosec^2θ + sin^2θ - 2cosecθsinθ
= 1 + 1 - 2cosecθsinθ
= 2 - 2cosecθsinθ
y^2 = (secθ – cosθ)^2
= sec^2θ + cos^2θ - 2secθcosθ
= 1 + 1 - 2secθcosθ
= 2 - 2secθcosθ
Step 2: Finding x^2y^2(x^2 + y^2 + 3)
x^2y^2(x^2 + y^2 + 3) = (2 - 2cosecθsinθ)(2 - 2secθcosθ)(2 - 2cosecθsinθ + 2 - 2secθcosθ + 3)
= 4(1 - cosecθsinθ)(1 - secθcosθ)(7 - 2cosecθsinθ - 2secθcosθ)
Now, 1 - cosecθsinθ = 1 - 1 = 0
1 - secθcosθ = 1 - 1 = 0
Therefore, the expression becomes:
4(0)(0)(7 - 0 - 0)
= 4(0)(7)
= 0
Therefore, x^2y^2(x^2 + y^2 + 3) = 0
Hence, the correct answer is option B - 0.
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If x = cosecθ – sinθ and y = secθ – cosθ, then the value of x2y2 (x2 + y2 + 3) is (SSC CGL 1st Sit. 2012)a)0b)1c)2d)3Correct answer is option 'B'. Can you explain this answer?
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