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If – π / 2 < x < π / 2, then the value of log sec x is

1) 2 cot h-1 (cosec2 (x / 2) - 1) 2) 2 cos h-1 (cosec2 (x / 2) - 1) 3) 2 cosec h-1 (cot2 (x / 2) - 1) 4) 2 cosec h-1 (cot2 (x / 2) + 1)... View Article

sin h^-1 (2^3/2) is

1) log (2 + √18) 2) log (3 + √8) 3) log (3 - √8) 4) log (√8 + √27) Solution: (2) log (3 + √8) sin h-1 (23/2) = log (23/2 + √(23/2)2 + 1 =... View Article

sec h^-1 (1 / 2) is

1) log (√3 + √2) 2) log (√3 + 1) 3) log (2 + √3) 4) None of these Solution: (3) log (2 + √3) sec h-1 (1 / 2) = cos h-1 (2) = log (2 + √22 -... View Article

log (3 + 2 √2) =

1) sin h-1 3 2) cos h-1 3 3) tan h-1 3 4) cot h-1 3 Solution: (2) cos h-1 3 log (3 + 2 √2) = log (3 + √8) = log (3 + √9 - 1) = log (3 + √32... View Article

If cos h^-1 x = log (2 + √3), then x =

1) 2 2) 1 3) 3 4) 5 Solution: (1) 2 cos h-1 x = log (2 + √3) cos h-1 x = log (x + √x2 - 1) = log (2 + √3) x = 2... View Article

If x = log (y + √y² + 1), then y =

1) tan h x 2) cot h x 3) sin h x 4) cos h x Solution: (3) sin h x x = log (y + √y2 + 1) = sin h-1 y y = sin h x... View Article

The period of sin h (x / 2) is

1) 2πi 2) cot h x 3) 4πi 4) cos h x Solution: (3) 4πi The period of sin h x is 2πi, hence, the period of sin h (x / 2) is 4πi.... View Article

The period of cos h (θ / 3) is

1) 6πi 2) 2πi 3) πi 4) 9πi Solution: (1) 6πi Since the period of coshθ is 2πi, the period of cos h (θ / 3) is 3 * 2πi = 6πi.... View Article

sec h (πi) + cosec h [(π / 2)i] =

1) 1 - i 2) - 1 + i 3) - 1 - i 4) 1 + i Solution: (3) - 1 - i sec h (πi) + cosec h [(π / 2)i] = sec π - i cosec (π / 2) = - 1 - i... View Article

sin h [(π / 6) i] =

1) - i / 2 2) i / 2 3) i √3 / 2 4) - i √3 / 2 Solution: (2) i / 2 sin h [(π / 6) i] = i * sin (π / 6) = i * (1 / 2) = i / 2... View Article

cos ix + i sin ix equals

1) eix 2) e-ix 3) ex 4) e-x Solution: (4) e-x cos ix + i sin ix = cos hx + i i sin hx = cos hx - sin hx = [ex + e-x - ex + e-x] / 2 = e-x... View Article

The value of sec h (iπ) is

1) -1 2) i 3) 0 4) 1 Solution: (1) -1 sec h (iπ) = 2 / [eπi + e-πi] = sec π = -1... View Article

If cos (x + iy) = A + iB, then A =

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

cos h2 + sin h2 =

1) 1 / e 2) e 3) 1 / e2 4) e2 Solution: (4) e2 cos h2 + sin h2 = {[e2 + e-2] / 2} + {[e2 - e-2] / 2} = e2... View Article

If tan h² x = tan² θ, then cosh 2x =

1) - sin 2θ 2) sec 2θ 3) cos 3θ 4) cos 2θ Solution: (2) sec 2θ cosh 2x = [1 + tanh2 x] / [1 - tanh2 x] = [1 + tan2 θ] / [1 - tan2 θ] = 1 /... View Article

[sin hx – sin hy] / [cos hx – cos hy] =

1) 2 cot {h (x + y)} 2) tan h {[x + y] / 2} 3) cot h {[x + y] / 2} 4) cot h {[x - y] / 2} Solution: (3) cot h {[x + y] / 2} [sin hx - sin... View Article

If sin^-1 2a / [1 + a²] – cos^-1 [1 – b²] / [1 + b²] = tan^-1 2x / [1 – x²], then x =

1) a 2) b 3) [a + b] / [1 - ab] 4) [a - b] / [1 + ab] Solution: (4) [a - b] / [1 + ab] Put a = tan θ, b = tan ɸ and x = tan ψ, then reduced... View Article

sin (2 sin^-1 0.8) =

1) 0.96 2) 0.48 3) 0.64 4) None of these Solution: (1) 0.96 sin (2 sin-1 0.8) = sin (2 sin-1 (4 / 5)) = sin (sin-1 2 * (4 / 5) * √1 - (16 /... View Article

If sec^-1 x = cosec^-1 y, then cos^-1 (1 / x) + cos^-1 (1 / y) =

1) π 2) π / 4 3) - π / 2 4) π / 2 Solution: (4) π / 2 sec-1 x = cosec-1 y cos-1 (1 / x) = sin-1 (1 / y) cos-1 (1 / x) = (π / 2) - cos-1 (1 /... View Article

If sin^-1 [x – (x² / 2) + (x³ / 4) – ….. + cos^-1 (x² – (x⁴ / 2) + (x⁶ / 4) – …. ) = π / 2. For 0 < |x| < √2, then x =

1) 1 / 2 2) 1 3) - 1 / 2 4) -1 Solution: (2) 1 sin-1 y + cos-1 y = π / 2, |y| ≤ 1 x - (x2 / 2) + (x3 / 4) - ...... = x2 - (x4 / 2) + (x6 /... View Article

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