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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

Find imaginary part of sin^-1 (cosec θ)

1) log [cot (θ / 2)] 2) π / 2 3) (1 / 2) log [cot (θ / 2)] 4) None of these Solution: (1) log [cot (θ / 2)] Let sin-1 (cosec θ) = x + iy... View Article

If cos (u + iv) = x + iy, then x² + y² + 1 =

1) cos2 u + sin h2 v 2) sin2 u + cos h2 v 3) cos2 u + cos h2 v 4) sin2 u + sin h2 v Solution: (3) cos2 u + cos h2 v cos (u + iv) = x + iy... View Article

If cos hɑ = sec ɑ, then tan² (x / 2) =

1) cos2 (ɑ / 2) 2) sin2 (ɑ / 2) 3) cot2 (ɑ / 2) 4) tan h2 (ɑ / 2) Solution: (4) tan h2 (ɑ / 2) cos h ɑ = sec x = 1 / cos x = 1 / [[1 - tan2... View Article

If u = log tan [(π / 4) + (x / 2)], then cos hu =

1) sec x 2) cosec x 3) tan x 4) sin x Solution: (1) sec x u = log tan [(π / 4) + (x / 2)] eu / 1 = [1 + tan (x / 2)] / [1 - tan (x / 2)]... View Article

sin h^-1 (1 / 2) =

1) tan h-1 √5 2) tan h-1 (1 / √5) 3) tan h-1 √3 4) tan h-1 (2 / √5) Solution: (2) tan h-1 (1 / √5) sin h-1 (1 / 2) = tan h-1 [(1 / 2) / [√1... View Article

If tan hx = 3 / 4, then the value of x is

1) √7 2) - √7 3) log √7 4) - log √7 Solution: (3) log √7 x = tan h-1 (3 / 4) = (1 / 2) log [(1 + (3 / 4)] / [1 - (3 / 4)]] = (1 / 2) log 7 =... View Article

2 tan h^-1 (1 / √2) =

1) cos h-1 (1 / 3) 2) cos h-1 √3 3) cos h-1 3 4) cos h-1 (√2 + 1) / (√2 - 1) Solution: (3) cos h-1 3 2 tan h-1 x = cos h-1 [1 + x2] / [1 -... View Article

cot h^-1 x =

1) (1 / 2) log [1 + x] / [1 - x] 2) (1 / 2) log [x + 1] / [x - 1] 3) (1 / 2) log [x - 1] / [x + 1] 4) None of these Solution: (2) (1 / 2) log... View Article

If sin^-1 (A + iB) = x + iy, then (A / B) =

1) tan x / tan hy 2) tan hx / tan y 3) tan hx / tan hy 4) cos x / cos hy Solution: (1) tan x / tan hy sin-1 (A + iB) = x + iy (A + iB) = sin... View Article

tan h^-1 x =

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

cos h^-1 x =

1) log [x + √x2 + 1] 2) log [x - √x2 - 1] 3) log [x - √x2 - 1] 4) log [x + √x2 - 1] Solution: (4) log [x + √x2 - 1] x = cos hy = [ey + e-y]... View Article

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