(i) ± (ii) x = 3 (iii) ± (iv) x =

Sol. (i) Let sin–1 x = θ

cos(2 θ ) = ⇒ 1 – 2 sin2 θ =

sin2 θ = ⇒ x = sin θ = ±

(ii) cot–1 x + tan–1 x = . so x = 3

(iii) tan − 1 + tan–1 = tan–1 = tan–1 =

case- (I)

= 1 ⇒ x = ± ... (1)

If ⇒ x ∈ (–2, 2) ...(2)

from (1) & (2) x = ±

case-(II) If

⇒ x ∈ (– ∞ , –2) ∪ (2, ∞ ) ....(3)

tan–1 + π =

= 1 x = ± .....(4)

from (3) & (4) ⇒ x ∈ φ

case-(III) ∴ = 1

no solution ⇒ x = ± are the solutions

(iv) sin–1 x = ⇒ sin x sin–1 x = sin

x = ⇒ x =

1 – 4x2 = 0 x = ±

But x = – does not satisfy

x =

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