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In a ∆PQR, P is the largest angle and cos P = 1/3, Further in the circle of the triangle touches the sides PQ, QR, and RP at N, L, and M respectively, such that the lengths of PN, QL, and RM are consecutive even integers, Then, possible length(s) of the side (s) of the triangle is (are)

1) 16 2) 18 3) 24 4) 22 Solution: Let s - a = 2k - 2 s - b = 2k s - c = 2k + 2, k ∈ I, k > 1 Adding we get, s = 6k So, a = 4k + 2 b =... View Article

If the sides of the triangles are p, q, √(p² + q² + pq), then the greatest angle is

1) π/2 2) 5π/4 3) 2π/3 4) 7π/4 Solution: Given p, q, √(p2 + q2 + pq) are the sides. cos θ = [p2 + q2 - √(p2 + q2 + pq)]/2pq = (p2 + q2 - p2... View Article

If in a ∆ABC, the altitudes from the vertices A, B, C on the opposite side are in HP, then sin A, sin B, and sin C are in

1) HP 2) Arithmetico - Geometric Progression 3) AP 4) GP Solution: Consider ∆BDA, cos (90 - B) = AD/c AD = c sin B Similarly, BE = a sin... View Article

If in the ∆ABC, ∠B = 45^o, then a⁴ + b⁴ + c⁴ is equal to

1) 2a2 (b2 + c2) 2) 2c2 (a2 + b2) 3) 2b2 (a2 + c2) 4) 2(a2b2 + b2c2 + c2a2) Solution: Given ∠B = 45o cos B = (a2 + c2 - b2)/2ac 1/√2 = (a2... View Article

In a ∆ABC, if the sides are a = 3, b = 5 and c = 4, then sin B/2 + cos B/2 is equal to

1) √2 2) (√3 + 1)/2 3) (√3 - 1)/2 4) 1 Solution: Given a = 3, b = 5 and c = 4 Here b2 = a2 + c2 So Pythagoras theorem is satisfied. Thus B... View Article

In a ∆ABC, a, c, A are given and b₁, b₂ are two values of the third side b such that b₂ = 2b₁, then sin A is equal to

1) √[(9a2 - c2 )/8a2] 2) √[(9a2 - c2 )/8c2] 3) √[(9a2 + c2 )/8c2] 4) None of these Solution: We know cos A = (b2 + c2 - a2)/2bc 2bc cos A =... View Article

The value of log tan [(π / 4) + (ix / 2)] is

1) i tan-1 (sin hx) 2) i tan-1 (sin hx) 3) i tan-1 (cos hx) 4) None of these Solution: (1) i tan-1 (sin hx) log tan [(π / 4) + (ix / 2)] =... View Article

2 sin h^-1 θ =

1) sin h-1 (2θ √1 + θ2) 2) sin h-1 (2θ √1 - θ2) 3) sin h-1 (θ √1 + θ2) 4) None of these Solution: (1) sin h-1 (2θ √1 + θ2) 2 sin-1 x =... View Article

If tan^-1 (ɑ + iβ) = x + iy, then x =

1) (1 / 2) tan-1 [(2ɑ) / (1 - ɑ2 - β2)] 2) (1 / 2) tan-1 [(2ɑ) / (1 + ɑ2 + β2)] 3) tan-1 [(2ɑ) / (1 - ɑ2 - β2)] 4) None of these Solution:... View Article

The imaginary part of tan^-1 (cos θ + i sin θ)

1) tan h-1 (sin θ) 2) tan h-1 (∞) 3) (1 / 2) tan h-1 (sin θ) 4) None of these Solution: (3) (1 / 2) tan h-1 (sin θ) tan-1 [cos θ + i sin θ]... View Article

The value of tan h^-1 (2^-1) is

1) log 2 2) log 2-1 3) log √3 4) None of these Solution: (3) log √3 tan h-1 x = (1 / 2) log {[1 + x] / [1 - x]} tan h-1 (1 / 2) = (1 / 2)... View Article

If cosec h^-1 1 = x + iy, then the value of y is

1) 1 2) 0 3) log (1 + √2) 4) -1 Solution: (2) 0 cosec h-1 x = log [1 + [(√1 + x2) / (x)]] cosec h-1 x = log [1 + √2] / 1 x + iy = log (√2 +... View Article

sec h^-1 (sin x) =

1) log cot (x / 2) 2) log tan (x / 2) 3) log cot x 4) None of these Solution: (1) log cot (x / 2) sec h-1 (sin x) = log [1 + (√1 - (sin... View Article

The value of 2 cot h^-1 (z / 2) is

1) log {[z - 1] / [z + 2]} 2) (1 / 2) log {[z - 1] / [z + 2]} 3) (1 / 2) log {[z + 1] / [z - 1]} 4) - log {[z - 2] / [z + 2]} Solution: (4) -... View Article

The imaginary part of sin² (x + iy) is

1) (1 / 2) cos h 2x cos 2y 2) (1 / 2) cos 2x cos h 2y 3) (1 / 2) sin h 2x sin 2y 4) (1 / 2) sin 2x sin h 2y Solution: (4) (1 / 2) sin 2x sin... View Article

tan h (πi / 4) – cot h (πi / 4) =

1) 0 2) 2 3) Root 2 4) None of these Solution: (4) None of these tan h (πi / 4) - cot h (πi / 4) = i tan (π / 4) + i cot (π / 4) = 2i... View Article

If sin x cos hy = cos θ and cos x sin hy = sin θ then sin h² y =

1) sin2 x 2) cos h2 x 3) cos2 x 4) 1 Solution: (3) cos2 x sin x cos hy = cos θ and cos x sin hy = sin θ Squaring and adding given relations... View Article

If tan (x / 2) cot h (x / 2) = 1, then the value of cos x cos hx is

1) 1 2) - 1 3) cos2 x 4) sin2 x Solution: (1) 1 tan (x / 2) = tan h (x / 2) tan (x / 2) = tan h (x / 2) [tan2 (x / 2] / 1 = [tan h2 (x / 2)]... View Article

If cosec θ = cot hx, then the value of tan θ is

1) cos hx 2) sin hx 3) tan hx 4) cosec hx Solution: (2) sin hx cosec θ = cot hx 1 + cot2 θ = cot h2 x cot2 θ = cot h2 x - 1 = cosec h2 x cot... View Article

If cos hz = sec θ, then sin hz =

1) cosec θ 2) cot θ 3) tan (θ / 2) 4) tan θ Solution: (4) tan θ cos h2 z - sin h2 z = 1 sin h2 z = cos h2 z - 1 sin h2 z = sec2 θ - 1 sin h2... View Article

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