Archives

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

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

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

sin h-1 x =

1) log [x + √1 - x2] 2) log [x + √x2 + 1] 3) log [x + √x2 - 1] 4) None of these Solution: (2) log [x + √x2 + 1] x = sin ht t = sin h-1 x... View Article

The period of cos h (4x) is

1) 2πi 2) πi 3) πi / 2 4) 2π Solution: (3) πi / 2 cos h (4x) = cos h (2πi + 4x) = cos h 4 [(πi / 2) + x] The period is πi / 2.... View Article

The period of cot h (nx / 4) is

1) πi / n 2) 4πi / n 3) nπi / 4 4) πi Solution: (2) 4πi / n cot h (nx / 4) = cot h [(πi + (nx / 4))] = cot h (n / 4) [4 * (πi / n) + x] The... View Article

sin2 ix + cos h2 x =

1) 1 2) -1 3) 2 cos h2 x 4) cos hx Solution: (1) 1 sin ix = i sin hx sin2 (ix) = (i sin hx)2 = - sin h2 x sin2 (ix) + cos h2 x = cos h2 x -... View Article