Find:\( \tan ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{4}\right) \)

1.	Find:\( \tan ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{4}\right) \)
Answer» Let \(\frac12\)cos^-1\(\frac34\) = θ ⇒ cos 2θ = 3/4 ⇒ 2 cos²θ - 1 = 3/4 ⇒ 2 cos²θ = 1 + 3/4 = 7/4 ⇒ cos²θ = 7/8 ⇒ cos θ = \(\frac{\sqrt7}{2\sqrt2}\) \(\therefore\) sin θ = \(\frac1{2\sqrt2}\) \(\therefore\) tan θ = \(\frac{sin\theta}{cos\theta}\) = \(\frac{1/2\sqrt2}{\sqrt 7/2\sqrt2}\) = \(\frac1{\sqrt7}\) \(\therefore\) θ = tan^-1\(\frac1{\sqrt7}\) \(\therefore\) θ = tan^-1\(\frac1{\sqrt7}\) ⇒ \(\frac12\) cos^-1\(\frac34\) = tan^-1\(\frac1{\sqrt7}\) \(\therefore\) tan²(\(\frac12\)cos^-1\(\frac34\)) = tan²(tan^-1\(\frac1{\sqrt7}\)) = [tan(tan^-1\(\frac1{\sqrt7}\))]² = (\(\frac1{\sqrt7}\))² = \(\frac17\)

Answer»

Let \(\frac12\)cos^-1\(\frac34\) = θ

⇒ cos 2θ = 3/4

⇒ 2 cos²θ - 1 = 3/4

⇒ 2 cos²θ = 1 + 3/4 = 7/4

⇒ cos²θ = 7/8

⇒ cos θ = \(\frac{\sqrt7}{2\sqrt2}\)

\(\therefore\) sin θ = \(\frac1{2\sqrt2}\)

\(\therefore\) tan θ = \(\frac{sin\theta}{cos\theta}\) = \(\frac{1/2\sqrt2}{\sqrt 7/2\sqrt2}\) = \(\frac1{\sqrt7}\)

\(\therefore\) θ = tan^-1\(\frac1{\sqrt7}\)

⇒ \(\frac12\) cos^-1\(\frac34\) = tan^-1\(\frac1{\sqrt7}\)

\(\therefore\) tan²(\(\frac12\)cos^-1\(\frac34\)) = tan²(tan^-1\(\frac1{\sqrt7}\)) = [tan(tan^-1\(\frac1{\sqrt7}\))]²

= (\(\frac1{\sqrt7}\))² = \(\frac17\)

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