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

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Question

(A) ધારો કે $x = \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}$.નોંધો કે $8-2 \sqrt{15} = (\sqrt{5}-\sqrt{3})^2$,તેથી $\sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqr

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) ધારો કે $x = \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}$.નોંધો કે $8-2 \sqrt{15} = (\sqrt{5}-\sqrt{3})^2$,તેથી $\sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqrt{3}$.આમ,$x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$.છેદનું સંમેયીકરણ કરતા: $x = \frac{(\sqrt{5}-\sqrt{3})^2}{5-3} = \frac{5+3-2 \sqrt{15}}{2} = \frac{8-2 \sqrt{15}}{2} = 4-\sqrt{15}$.આપણે જાણીએ છીએ કે $\operatorname{Tan}^{-1}(4-\sqrt{15}) = \operatorname{Tan}^{-1} \left( \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \right) = \operatorname{Tan}^{-1} \left( \frac{\sqrt{5}}{\sqrt{3}} \right) - \operatorname{Tan}^{-1}(1) = \operatorname{Tan}^{-1} \left( \sqrt{\frac{5}{3}} \right) - \frac{\pi}{4}$.વૈકલ્પિક રીતે,$\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} \right)$ નો ઉપયોગ કરીને,પદાવલિ $\operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$ માં પરિણમે છે.

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

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