Question

tan^-1(2 cos(2sin^-1 1÷2) . Find the value . No spam please.​

Answer 1

Answer:

$\mathrm{\dfrac{\pi}{3}}$

Step-by-step explanation:

Given :

$\mathrm{tan^{-1}\bigg(2cos\bigg(sin^{-1}\bigg(\dfrac{1}{2}\bigg)\bigg)\bigg)}$

To find :

The value of the given expression

Solution :

$\longmapsto \mathrm{tan^{-1}\bigg(2cos\bigg(sin^{-1}\bigg(\dfrac{1}{2}\bigg)\bigg)\bigg)}$

We know that,

$\boxed{\mathrm{sin^{-1}\bigg(\dfrac{1}{2}\bigg) = \dfrac{\pi}{6}}}$

$\implies \mathrm{tan^{-1}\bigg(2cos\bigg(\dfrac{\pi}{6}\bigg)\bigg)}$

$\implies \mathrm{tan^{-1}\bigg(2\times \dfrac{\sqrt3}{2}\bigg)}$

$\implies \mathrm{tan^{-1}\bigg(\not2\times \dfrac{\sqrt3}{\not2}\bigg)}$

$\implies \mathrm{tan^{-1}\sqrt3}$

We know that,

$\boxed{\mathrm{tan^{-1}\sqrt3 = \dfrac{\pi}{3}}}$

$\implies \mathrm{\underline{\underline{\dfrac{\pi}{3}}}}$

$\therefore \underline{\boxed{\mathbf{tan^{-1}\bigg(2cos\bigg(sin^{-1}\bigg(\dfrac{1}{2}\bigg)\bigg)\bigg) = \dfrac{\pi}{3}}}}$

❖ Extra information :

⟡ Trigonometric Values table :

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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