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Two sides of a triangle are given by the roots of the equations x² – 5x + 6 = 0 and the angle between the sides is π/3. Then, the perimeter of the triangle is

1) 5 + √2 2) 5 + √3 3) 5 + √5 4) 5 + √7 Solution: Given x2 - 5x + 6 = 0 (x - 2)(x - 3) = 0 So x = 2 and x = 3 Let b = 2 and c = 3 Angle... View Article

In ∆ABC, if (a + b + c) (a – b + c) = 3ac, then

1) ∠B = 60o 2) ∠B = 30o 3) ∠C = 60o 4) ∠A + ∠C = 90o Solution: Given (a + b + c) (a - b + c) = 3ac ((a + c) + b)((a + c) - b) = 3ac (a + c)2... View Article

If a₁, a₂,….. a_n is an AP with common difference d, then tan [tan^-1 d / [1 + a₁a₂] + tan^-1 d / [1 + a₂a₃] + …. + tan^-1 d / [1 + a_n-1a_n]] =

1) [n - 1] d / [a1 + an] 2) [n - 1] d / [1 + a1an] 3) nd / [1 + a1an] 4) [an - a1] / [an + a1] Solution: (2) [n - 1] d / [1 + a1an] tan-1 d... View Article

If tan^-1 x + cos^-1 y / (√1 + y²) = sin^-1 3 / √10 and both x and y are positive and integral, then x and y =

1) (1, 2) and (2, 7) 2) (1, 2) and (1, 7) 3) (1, 7) and (2, 7) 4) (1, 7) and (2, 1) Solution: (1) (1, 2) and (2, 7) tan-1 x + tan-1 (1 / y)... View Article

If x₁, x₂, x₃, x₄ are roots of the equation x⁴ – x³ sin 2β + x² cos 2β – x cos β – sin β = 0, the tan^-1 x₁ + tan^-1 x₂ + tan^-1 x₃ + tan^-1 x₄ =

1) β 2) (π / 2) - β 3) π - β 4) - β Solution: (2) (π / 2) - β ∑x1 = sin 2β, ∑x1x2 = cos 2β, ∑x1x2x3 = cos β and x1x2x3x4 = - sin β tan-1 x1... View Article

If cos^-1 (x / a) + cos^-1 (y / b) = ɑ, then (x² / a²) – (2xy / ab) cos ɑ + (y² / b²) =

1) sin2 ɑ 2) cos2 ɑ 3) tan2 ɑ 4) cot2 ɑ Solution: (1) sin2 ɑ cos-1 [(x / a) * (y / b) - √1 - (x2 / a2) √1 - (y2 / b2)] = ɑ (xy / ab) - √1 -... View Article

If a, b, c be positive real number and the value of θ = tan^-1 √a (a + b + c) / bc + tan^-1 √b (a + b + c) / ca + tan^-1 √c (a + b + c) / ab, then tan θ =

1) 0 2) 1 3) [a + b + c] / abc 4) None of these Solution: (1) 0 θ = tan-1 √a (a + b + c) / bc + tan-1 √b (a + b + c) / ca + tan-1 √c (a + b... View Article

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Archives

Two sides of a triangle are given by the roots of the equations x² – 5x + 6 = 0 and the angle between the sides is π/3. Then, the perimeter of the triangle is

In ∆ABC, if (a + b + c) (a – b + c) = 3ac, then

If a₁, a₂,….. a_n is an AP with common difference d, then tan [tan^-1 d / [1 + a₁a₂] + tan^-1 d / [1 + a₂a₃] + …. + tan^-1 d / [1 + a_n-1a_n]] =

If tan^-1 x + cos^-1 y / (√1 + y²) = sin^-1 3 / √10 and both x and y are positive and integral, then x and y =

If x₁, x₂, x₃, x₄ are roots of the equation x⁴ – x³ sin 2β + x² cos 2β – x cos β – sin β = 0, the tan^-1 x₁ + tan^-1 x₂ + tan^-1 x₃ + tan^-1 x₄ =

If cos^-1 (x / a) + cos^-1 (y / b) = ɑ, then (x² / a²) – (2xy / ab) cos ɑ + (y² / b²) =

If a, b, c be positive real number and the value of θ = tan^-1 √a (a + b + c) / bc + tan^-1 √b (a + b + c) / ca + tan^-1 √c (a + b + c) / ab, then tan θ =

If in a triangle ABC, A = tan^-1 2 and B = tan^-1 3, then angle C =

Considering only the principal values, if tan (cos^-1 x) = sin [cot^-1 (1 / 2)], then x =

If x + (1 / x) = 2, the principal value of sin^-1 x is

If sin^-1 x + sin^-1 y + sin^-1 z = 3π / 2, then the value of x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ – 9 / [x¹⁰¹ + y¹⁰¹ + z¹⁰¹] =

If θ = sin^-1 x + cos^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which θ lies is given by

If tan^-1 x + tan^-1 y + tan^-1 z = π, then x + y + z =

The equation 2 cos^-1 x + sin^-1 x = 11π / 6 has

∑_m=1ⁿ tan^-1 [2m] / [m⁴ + m² + 2] =

tan [(π / 4) + (1 / 2) cos^-1 (a / b)] + tan^-1 [(π / 4) – (1 / 2) cos^-1 (a / b)] =

tan^-1 [√1 + x² + √1 – x²] / [√1 + x² – √1 – x²]

2 tan^-1 [√(a – b / a + b) tan (θ / 2)] =

If θ = tan^-1 a, ɸ = tan^-1 b and ab = – 1, then θ – ɸ =

2 tan^-1 (cos x) = tan^-1 (cosec² x), then x =

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