Question

Cot 60 degree minus Cos 45 degree

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{cot\,60^\circ-cos\,45^\circ}$

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

$\textsf{The value of}\;\mathsf{cot\,60^\circ-cos\,45^\circ}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{cot\,60^\circ-cos\,45^\circ}$

$\textsf{This can be written as}$

$\mathsf{=\dfrac{1}{tan\,60^\circ}-cos\,45^\circ}$

$\mathsf{=\dfrac{1}{\sqrt3}-\dfrac{1}{\sqrt2}}$

$\mathsf{=\dfrac{\sqrt2-\sqrt3}{\sqrt3\sqrt2}}$

$\mathsf{=\dfrac{\sqrt2-\sqrt3}{\sqrt6}}$

$\implies\boxed{\bf\,cot\,60^\circ-cos\,45^\circ=\dfrac{\sqrt2-\sqrt3}{\sqrt6}}$

$\textbf{Standard trigonometric table:}$

$\left\begin{array}{|c|c|c|c|c|c|}\cline{1-6}&0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\\\cline{1-6}\bf\,sin\theta&0&\frac{1}{2}&\frac{1}{\sqrt2}&\frac{\sqrt3}{2}&1\\\cline{1-6}\bf\,cos\theta&1&\frac{\sqrt3}{2}&\frac{1}{\sqrt2}&\frac{1}{2}&0\\\cline{1-6}\bf\,tan\theta&0&\frac{1}{\sqrt3}&1&\sqrt3&\infty\\\cline{1-6}\end{array}\right$

Answer 2

Step-by-step explanation:

$\sf We \: have,$

$\sf \cot 60^{\circ} - \cos 45^{\circ}$

$\sf : \implies \bf{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{2}}}$

$\sf : \implies \bf{\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{3}\sqrt{2}}}$

$\sf : \implies\underline{\boxed{\bf{\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}}}}}$