If $ 3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}+2 \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3} $, then the value of $ x $ lying in $ \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $ is

Question

If $ 3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}+2 \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3} $, then the value of $ x $ lying in $ \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $ is, $ \frac{1}{2} $, $ \frac{1}{\sqrt{3}} $, $ \frac{\sqrt{3}}{2} $, $ -\frac{\sqrt{3}}{2} $

MARKS App · Accepted Answer

3 sin - 1 ⁡ 2 x 1 + x 2 - 4 cos - 1 ⁡ 1 - x 2 1 + x 2 + 2 tan - 1 ⁡ 2 x 1 - x 2 = π 3 On putting x = tan ⁡ θ , we get 3 sin - 1 ⁡ 2 tan ⁡ θ 1 + tan 2 ⁡ θ - 4 cos - 1 ⁡ 1 - tan 2 ⁡ θ 1 + tan 2 ⁡ θ + 2 tan - 1 ⁡ 2 tan ⁡ θ 1 - tan 2 ⁡ θ = π 3 ⇒ 3 sin - 1 ⁡ ( sin ⁡ 2 θ ) - 4 cos - 1 ⁡ cos ⁡ 2 θ + 2 tan - 1 ⁡ ( tan ⁡ 2 θ ) = π 3 ⇒ 3 2 θ - 4 2 θ + 2 2 θ = π 3 ⇒ 6 θ - 8 θ + 4 θ = π 3 ⇒ θ = π 6 ⇒ tan - 1 ⁡ x = π 6 ⇒ x = tan ⁡ π 6 ⇒ x = 1 3

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