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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

If we consider only the principal values of the inverse trigonometric functions, then the value of tan [cos^-1 (1 / 5√2) – sin^-1 (4 / √17)] is

1) √29 / 3 2) 29 / 3 3) √3 / 29 4) 3 / 29 Solution: (4) 3 / 29 tan [cos-1 (1 / 5√2) - sin-1 (4 / √17)] = tan [tan-1 7 - tan-1 4] = tan... View Article

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fences are of same length x. The maximum area enclosed by the park is

1) √(x3/8) 2) x2/2 3) πx2 4) 3x2/2 Solution: Let ABC be the triangle such that AB = AC = x And let ∠ACB = sin θ Draw AD perpendicular to BC.... View Article

Points D, and E are taken on the side BC of the ∆ABC, such that BD = DE = EC. If ∠BAD = x, ∠DAE = y and ∠EAC = z, then the value of sin(x + y) sin (y + z)/sin x sin z is

1) 1 2) 2 3) 4 4) None of these Solution: Consider the figure below. From ∆ADC, sin (y + z)/DC = sin C/AD .....(1) From ∆ABD sin x/BD =... View Article

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Archives

sin h [(π / 6) i] =

cos ix + i sin ix equals

The value of sec h (iπ) is

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

cos h2 + sin h2 =

If tan h² x = tan² θ, then cosh 2x =

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

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

sin (2 sin^-1 0.8) =

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

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

If we consider only the principal values of the inverse trigonometric functions, then the value of tan [cos^-1 (1 / 5√2) – sin^-1 (4 / √17)] is

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fences are of same length x. The maximum area enclosed by the park is

Points D, and E are taken on the side BC of the ∆ABC, such that BD = DE = EC. If ∠BAD = x, ∠DAE = y and ∠EAC = z, then the value of sin(x + y) sin (y + z)/sin x sin z is

The sides of a triangle are in the ratio 1 : √3 : 2, then the angles of the triangle are in the ratio

In a ∆ABC, if tan A/2 = 5/6, tan C/2 = 2/5, then

If in ∆ABC, sin A/2 sin C/2 = sin B/2 and 2s is the perimeter of the triangle, then s is equal to

In a ∆ABC , (b + c) cos A + (a + c) cos B + (a + b) cos C is equal to

In ∆ABC, if cot A, cot B, and cot C are in AP, then a², b², and c²are in

If in a ∆ABC, 4 sin A = 4 sin B = 3 sin C, the cos C is equal to

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