Question

Sec inverse 2 root 2÷1+root 3

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{sec^{-1}(\dfrac{2\sqrt{2}}{1+\sqrt{3}})}$

$\underline{\textsf{To find:}}$

$\mathsf{The value of }\;\mathsf{sec^{-1}(\dfrac{2\sqrt{2}}{1+\sqrt{3}})}$

$\underline{\textsf{Solution:}}$

$\textsf{Take,}$

$\mathsf{sec^{-1}(\dfrac{2\sqrt{2}}{1+\sqrt{3}})=\theta}$

$\mathsf{\dfrac{2\sqrt{2}}{1+\sqrt{3}}=sec\,\theta}$

$\textsf{Taking reciprocals, we get}$

$\mathsf{cos\theta=\dfrac{1+\sqrt{3}}{2\sqrt{2}}}$

$\mathsf{cos\theta=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}}$

$\mathsf{cos\theta=\dfrac{1}{2}\;\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\;\dfrac{1}{\sqrt{2}}}$

$\mathsf{cos\theta=cos60^{\circ}\,cos45^{\circ}+sin60^{\circ}\,sin45^{\circ}}$

$\textsf{Using}$

$\boxed{\mathsf{cos(A-B)=cosA\,cosB+sinA\,sinB}}$

$\mathsf{cos\theta=cos(60^{\circ}-45^{\circ})}$

$\mathsf{cos\theta=cos\,15^{\circ}}$

$\implies\mathsf{\theta=15^{\circ}}$

$\therefore\textsf{The value of }\;\mathsf{sec^{-1}(\dfrac{2\sqrt{2}}{1+\sqrt{3}})=15^{\circ}}$