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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: cot12° cot38° cot52° cot60° cot78° = 1/√3

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This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to prove that ...Read more

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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: tan5° tan25° Tan30° tan65° tan85° = 1/√3

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This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to show that ...Read more

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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: cotθ tan(90°-θ) – sec(90°-θ) cosecθ + √3tan12° tan60° tan78° = 2

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This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to prove that ...Read more

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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: [sec(90°-θ) cosecθ – tan(90°-θ) cotθ + cos²25° + cos²65°]/3tan27°tan63° = 2/3

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This question is from trigonometry topic – trigonometric ratios on complementary angles in which we have been asked to prove ...Read more

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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: cos(90°-θ)/1+sin(90°-θ) + 1+sin(90°-θ)/cos(90°-θ) = 2cosecθ

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This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to prove that ...Read more

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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: [cos(90°-θ) sec(90°-θ) tan θ]/[cosec(90°-θ) sin(90°-θ) cot(90°-θ] tan(90°-θ)/cotθ = 2

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This question is from trigonometry topic – trigonometric ratios on complementary angles in which we have to show that [cos(90°-θ) ...Read more

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Rajan@2021Guru

Asked: April 8, 2021In: CBSE

Prove that: sinθ cos(90°-θ) cosθ/sin(90°-θ) + cosθ sin(90°-θ) sinθ/cos(90°-θ) = 1

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This question is from trigonometry topic – trigonometric ratios on complementary angles in which we have been asked to prove ...Read more

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Prove that: cot12° cot38° cot52° cot60° cot78° = 1/√3

Prove that: tan5° tan25° Tan30° tan65° tan85° = 1/√3

Prove that: cotθ tan(90°-θ) – sec(90°-θ) cosecθ + √3tan12° tan60° tan78° = 2

Prove that: [sec(90°-θ) cosecθ – tan(90°-θ) cotθ + cos²25° + cos²65°]/3tan27°tan63° = 2/3

Prove that: cos(90°-θ)/1+sin(90°-θ) + 1+sin(90°-θ)/cos(90°-θ) = 2cosecθ

Prove that: [cos(90°-θ) sec(90°-θ) tan θ]/[cosec(90°-θ) sin(90°-θ) cot(90°-θ] tan(90°-θ)/cotθ = 2

Prove that: sinθ cos(90°-θ) cosθ/sin(90°-θ) + cosθ sin(90°-θ) sinθ/cos(90°-θ) = 1