Trigonometry
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The value of sec 
cos²A(sin A + cosA) + sin² A(sin A - cos A) 
(sec² A - cosec² A) cosec²A(sin A - cos A) sec² A(sin A + coa A)
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Expression
cos²A(sinA + cosA) + sin²A(sinA - cosA) 
× 
1 - 1 cosec²A(sinA - cosA) sec²A(sinA + cosA) cos²A sin²A cos²A.sin²A (sinA + cosA) + sin²A.cos²A(sinA - cosA) × sin²A - cos²A sinA - cosA (sinA + cosA) sin²A.cos²A = sinA + cosA + sinA - cosA (sin²A - cos²A) sinA - cosA sinA + cosA (sinA + cosA)² + (sinA - cosA)² (sin²A - cos²A) (sinA - cosA)(sinA + cosA)
= 2(sin²A + cos²A) = 2Correct Option: C
Expression
cos²A(sinA + cosA) + sin²A(sinA - cosA) × 1 - 1 cosec²A(sinA - cosA) sec²A(sinA + cosA) cos²A sin²A cos²A.sin²A (sinA + cosA) + sin²A.cos²A(sinA - cosA) × sin²A - cos²A sinA - cosA (sinA + cosA) sin²A.cos²A = sinA + cosA + sinA - cosA (sin²A - cos²A) sinA - cosA sinA + cosA (sinA + cosA)² + (sinA - cosA)² (sin²A - cos²A) (sinA - cosA)(sinA + cosA)
= 2(sin²A + cos²A) = 2
- The eliminant of q from x cosθ – y sin θ = 2 and x sin θ + y cos θ = 4 will give
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x cos θ – y sin θ = 2
x sin θ + y cos θ = 4
On squaring both the equations and adding
x²cos²θ + y²sin²θ – 2 xy sin θ . cos θ + x²sin²θ + y² cos²θ + 2xy sin θ . cos θ
= 4 + 16
⇒ x² (cos²θ + sin²θ ) + y²
(sin²θ + cos²θ ) = 20
⇒ x² + y² = 20Correct Option: A
x cos θ – y sin θ = 2
x sin θ + y cos θ = 4
On squaring both the equations and adding
x²cos²θ + y²sin²θ – 2 xy sin θ . cos θ + x²sin²θ + y² cos²θ + 2xy sin θ . cos θ
= 4 + 16
⇒ x² (cos²θ + sin²θ ) + y²
(sin²θ + cos²θ ) = 20
⇒ x² + y² = 20
- If tan θ + cot θ = 2, then the value of tan2θ + cot2θ is
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tan θ + cot θ = 2
On squaring both sides,
(tan θ + cot θ )² = 4
⇒ tan²θ + cot²θ + 2tan θ . cot θ = 4
⇒ tan²θ + cot²θ = 4 – 2 = 2
[tan θ .cot θ = 1]Correct Option: A
tan θ + cot θ = 2
On squaring both sides,
(tan θ + cot θ )² = 4
⇒ tan²θ + cot²θ + 2tan θ . cot θ = 4
⇒ tan²θ + cot²θ = 4 – 2 = 2
[tan θ .cot θ = 1]
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the value of sec θ 1 + sin θ + cos θ - 2tan²θ is cos θ 1 + sin θ
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Expression
= sec θ 1 + sin θ + cos θ - 2 tan²θ cos θ 1 + sin θ = 1 + sin² θ + 2sin θ + cos² θ - 2 tan² θ cos² θ(1 + sin θ) = 2 - 2 tan² θ cos² θ
= 2 sec² θ - 2 tan² θ
=2(sec² θ - tan² θ) = 2Correct Option: C
Expression
= sec θ 1 + sin θ + cos θ - 2 tan²θ cos θ 1 + sin θ = 1 + sin² θ + 2sin θ + cos² θ - 2 tan² θ cos² θ(1 + sin θ) = 2 - 2 tan² θ cos² θ
= 2 sec² θ - 2 tan² θ
=2(sec² θ - tan² θ) = 2
- The value of 3 (sin x – cos x)4 + 6 (sin x + cos x)2+ 4 (sin6 x + cos6 x) is
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3 (sin x – cos x)4 + 6 (sinx + cos x)2 + 4 (sin6x + cos6 x)
= 3 (sin2x + cos2x – 2 sinx . cosx)2 + 6(sin2x + cos2 x + 2 sinx . cosx ) + 4[(sin2x + cos2x)3 – 3 sin2x . cos2x (sin2x + cos2x)]
= 3 (1 – 2 sinx cosx)² + 6(1 + 2 sinx . cosx) + 4(1 – 3 sin²x cos²x)
= 3 (1 + sin²x . cos²x – 4 sinx cosx) + 6 (1+ 2 sinx cosx) + 4 (1 – 3 sin²x cos²x)
= 3 + 6 + 4 = 13Correct Option: D
3 (sin x – cos x)4 + 6 (sinx + cos x)2 + 4 (sin6x + cos6 x)
= 3 (sin2x + cos2x – 2 sinx . cosx)2 + 6(sin2x + cos2 x + 2 sinx . cosx ) + 4[(sin2x + cos2x)3 – 3 sin2x . cos2x (sin2x + cos2x)]
= 3 (1 – 2 sinx cosx)² + 6(1 + 2 sinx . cosx) + 4(1 – 3 sin²x cos²x)
= 3 (1 + sin²x . cos²x – 4 sinx cosx) + 6 (1+ 2 sinx cosx) + 4 (1 – 3 sin²x cos²x)
= 3 + 6 + 4 = 13
