operatorname{Tan}^{-1} left( frac{sqrt{8-2 sq | vedclass

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Question

(A) Let $x = \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}$.Note that $8-2 \sqrt{15} = (\sqrt{5}-\sqrt{3})^2$,so $\sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqrt{3}$

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) Let $x = \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}$.Note that $8-2 \sqrt{15} = (\sqrt{5}-\sqrt{3})^2$,so $\sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqrt{3}$.Thus,$x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$.Rationalizing the denominator: $x = \frac{(\sqrt{5}-\sqrt{3})^2}{5-3} = \frac{5+3-2 \sqrt{15}}{2} = \frac{8-2 \sqrt{15}}{2} = 4-\sqrt{15}$.We know that $\operatorname{Tan}^{-1}(4-\sqrt{15}) = \operatorname{Tan}^{-1} \left( \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \right) = \operatorname{Tan}^{-1} \left( \frac{\frac{\sqrt{5}}{\sqrt{3}}-1}{\frac{\sqrt{5}}{\sqrt{3}}+1} \right) = \operatorname{Tan}^{-1} \left( \frac{\sqrt{5}}{\sqrt{3}} \right) - \operatorname{Tan}^{-1}(1) = \operatorname{Tan}^{-1} \left( \sqrt{\frac{5}{3}} \right) - \frac{\pi}{4}$.Alternatively,using $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} \right)$,the expression simplifies to $\operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$.

$\operatorname{Tan}^{-1} \left( \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1} \right) + \operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{5}} \right) =$

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