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📐 Mathematics  ·  Inverse Trigonometric Functions  ·  JEE

Evaluate tan-1(1) + tan-1(2) + tan-1(3).

Answer: pi.

  • A pi
  • B pi/2
  • C 3pi/4
  • D 2pi/3

Correct answer: A. pi

Explanation: tan-1(2)+tan-1(3) = pi + tan-1[(2+3)/(1-6)] = pi + tan-1(-1) = pi - pi/4 = 3pi/4 (since xy=6>1 and both positive, add pi). Then tan-1(1) + 3pi/4 = pi/4 + 3pi/4 = pi.

y = sin⁻¹x: Principal Value Branchxyx=-1x=1y=-π/2y=π/2Domain restricted to [-1,1]; range restricted to [-π/2,π/2] - this restriction is what makes the inverse exist

The graph of sin⁻¹x is confined to a narrow domain [-1,1] (since sine itself only takes values in that range) and range [-π/2,π/2] (the principal value branch chosen to make sine one-one and invertible there).

Concept context

Restricting trig functions to make them invertible, the principal value branches of sin-inverse, cos-inverse, tan-inverse, and friends, and the key identities relating them.

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