Question

What is the value of :

sin^3 60 cot 30 - 2 sec^2 45 + 3 cos 60 tan 45 - tan^2 60

Answer 1

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

$\mathsf{sin^360^\circ\;cot\,30^\circ-2\;sec^245^\circ+3\;cos\,60^\circ\;tan\,45^\circ-tan^260^\circ}$

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

$\textsf{The value of}$

$\mathsf{sin^360^\circ\;cot\,30^\circ-2\;sec^245^\circ+3\;cos\,60^\circ\;tan\,45^\circ-tan^260^\circ}$

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

$\mathsf{Consider,}$

$\mathsf{sin^360^\circ\;cot\,30^\circ-2\;sec^245^\circ+3\;cos\,60^\circ\;tan\,45^\circ-tan^260^\circ}$

$\mathsf{=\left(\dfrac{\sqrt3}{2}\right)^3\sqrt3-2\;(\sqrt2)^2+3\left(\dfrac{1}{2}\right)(1)-(\sqrt3)^2}$

$\mathsf{=\left(\dfrac{3\sqrt3}{8}\right)\sqrt3-2\;(2)+\dfrac{3}{2}-3}$

$\mathsf{=\dfrac{9}{8}-4+\dfrac{3}{2}-3}$

$\mathsf{=\dfrac{9}{8}+\dfrac{3}{2}-7}$

$\mathsf{=\dfrac{9+12-56}{8}}$

$\mathsf{=\dfrac{21-56}{8}}$

$\mathsf{=\dfrac{-35}{8}}$

$\implies\boxed{\mathsf{sin^360^\circ\;cot\,30^\circ-2\;sec^245^\circ+3\;cos\,60^\circ\;tan\,45^\circ-tan^260^\circ=\dfrac{-35}{8}}}$

$\textbf{Standard trigonometric table:}$

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

Answer 2

Answer:

$\frac{-35}{8}$

Step-by-step explanation:

Given:-  $sin^{3}$60° cot 30° - 2 $sec^{2}$ 45° + 3 cos 60° tan 45° - $tan^{2}$ 60°

To Find:- The value of the above equation.

Solution:-

=  $sin^{3}$60° cot 30° - 2 $sec^{2}$ 45° + 3 cos 60° tan 45° - $tan^{2}$ 60°

= $(\frac{\sqrt{3} }{2} )^{3}$ $\sqrt{3}$ - 2 $(\sqrt{2} )^{2}$ + 3 $(\frac{1}{2})$ (1) - $(\sqrt{3} )^{2}$

= $\frac{9}{8}$ - 4 + $\frac{3}{2}$ - 3

= $\frac{9 + 12 - 56}{8}$

= $\frac{-35}{8}$

#SPJ2