→ (p), → (p), → (p), → (s)

Sol.

Let x = , y = , z = , x, y, z > 0

θ = tan –1 x + tan –1 y + tan –1 z

Now x + y + z = + + =

and , xyz = ⇒ x + y + z = xyz ⇒ tan –1 x + tan –1 y + tan –1 z = π

Hence θ = π

Let α = tan –1 (cotA) ⇒ β = tan –1 (cot 3 A) ⇒ tan( α + β ) =

R.H.S. is negative ⇒

⇒ tan ( α + β – π ) = = – ⇒ α + β = π – tan –1

G.E. = π independent of A.

If x < 0, then {cos – 1 (2x 2 – 1) + 2cos –1 x} ⇒ x = cos θ , π/2 < θ < π

{cos – 1 (2x 2 – 1) + 2cos –1 x} = {cos – 1 (cos2 θ ) + 2cos –1 x}

= {cos – 1 (cos2 θ ) + 2cos –1 x} = {–2 θ + 2 π + 2 θ } = π

sin –1 – cos –1 + cos –1 = sin –1 – sin –1 + cos –1

= sin –1 + cos –1

= sin –1 + cos –1 = sin –1 + cos –1 =

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