If r2 = x2 + y2 + z2 and tan-1yzxr + tan-1xzyr&nbs

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 Multiple Choice QuestionsMultiple Choice Questions

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91.

If r2 = x2 + y2 + z2 and tan-1yzxr + tan-1xzyr = π2 - tan-1ϕ, then

  • ϕ = zrxy

  • ϕ = xyzr

  • ϕ = x + yzr

  • ϕ = yzxr + xzyr


B.

ϕ = xyzr

We have, tan-1yzxr + tan-1xzyr = π2 - tan-1ϕNow, tan-1yzxr + xzyr1 - yzxrxzyr = π2 - tan-1ϕ              tan-1x + tan-1y = tan-1x + y1 - xy tan-1y2zr + x2zrxyr2 - xyz2 = π2 - tan-1ϕ    tan-1zry2 + x2xyr2 - z2 = π2 - tan-1ϕ  tan-1zr x2 + y2xyx2 + y2 = π2 - tan-1ϕ               x2 + y2 = r2 - z2 tan-1zrxy = cot-1ϕ                tan-1ϕ + cot-1ϕ = π2 tan-1zrxy = tan-11ϕ                tan-11θ = cot-1θ              1ϕ = zrxy               ϕ = xyzr


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92.

If α = sin-1cossin-1x and β = cos-1sincos-1x, then tanα . tanβ is equal to

  • 1

  • - 1

  • 2

  • 12


93.

The value of 2tan-1csctan-1x - tancot-1x is

  • tan-1x

  • tan(x)

  • cot(x)

  • csc-1x


94.

If tan-1x2 + cot-1x2 = 5π28, then x is equal to

  • - 1

  • 1

  • 0

  • None of these


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95.

α32csc212tan-1αβ + β32sec21212tan-1βα is equal to

  • α - βα2 + β2

  • α + βα2 - β2

  • α + βα2 + β2

  • None of the above


96.

tanπ4 + 12cos-1ab + tanπ4 - 12cos-1ab is equal to

  • 2ab

  • 2ba

  • ab

  • ba


97.

The solution of equation 

sin-11 - x2 - 2sin-1x2 = π2 is

  • x = 0

  • x = 12

  • x = 0 and 12

  • None of these


98.

The value of sinπ2 - sin-1- 32 is

  • 12

  • - 12

  • 1

  • - 1


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99.

If a  sin-1x + cos-1x + tan-1x  b, then

  • a = 0, b = π

  • a = 0, b = π2

  • a = π2, b = π

  • None of these


100.

If sin-1x + sin-1y = 2π3 and cos-1x + cos-1y = π3. Then, (x, y) is equal to

  • (0, 1)

  • (1/2, 1)

  • (1, 1/2)

  • 3/2, 1


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