What does $- 2 cos (arctan (6)) + csc (a r c sec (2))$ $-2cos(arctan(6))+csc(arcsec(2))$ equal?

Question

What does $- 2 cos (arctan (6)) + csc (a r c sec (2))$ $-2cos(arctan(6))+csc(arcsec(2))$ equal?

1 Answer

Impact of this question

1831 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-18T22:37:49

$- 2 cos (arctan (6)) + csc (arcsec (2))$ $-2cos(arctan(6))+csc("arcsec"(2))$

$= - \frac{2}{\sqrt{37}} + \frac{2}{\sqrt{3}}$ $=-2/sqrt(37)+2/sqrt(3)$

$= \frac{- 6 \sqrt{37} + 74 \sqrt{3}}{111}$ $=(-6sqrt(37)+74sqrt(3))/111$

Explanation:

$arctan (6)$ $arctan(6)$ is one of the angles in a right angled triangle with legs of length $1$ $1$ , $6$ $6$ and hypotenuse $\sqrt{1^{2} + 6^{2}} = \sqrt{37}$ $sqrt(1^2+6^2) = sqrt(37)$

Hence $cos (arctan (6)) = \frac{1}{\sqrt{37}}$ $cos(arctan(6)) = 1/sqrt(37)$

$arcsec (2)$ $"arcsec"(2)$ is one of the angles in a right angled triangle with legs of length $1$ $1$ , $\sqrt{2^{2} - 1^{2}} = \sqrt{3}$ $sqrt(2^2-1^2) = sqrt(3)$ and hypotenuse $2$ $2$ .

Hence $csc (arcsec (2)) = \frac{2}{\sqrt{3}}$ $csc("arcsec"(2)) = 2/sqrt(3)$

So:

$- 2 cos (arctan (6)) + csc (arcsec (2))$ $-2cos(arctan(6))+csc("arcsec"(2))$

$= - \frac{2}{\sqrt{37}} + \frac{2}{\sqrt{3}}$ $=-2/sqrt(37)+2/sqrt(3)$

$= - \frac{2 \sqrt{37}}{37} + \frac{2 \sqrt{3}}{3}$ $=-(2sqrt(37))/37+(2sqrt(3))/3$

$= \frac{- 6 \sqrt{37} + 74 \sqrt{3}}{111}$ $=(-6sqrt(37)+74sqrt(3))/111$

Science

Math

Humanities

... and beyond

What does $- 2 cos (arctan (6)) + csc (a r c sec (2))$ $-2cos(arctan(6))+csc(arcsec(2))$ equal?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What does −2cos(arctan(6))+csc(arcsec(2))-2cos(arctan(6))+csc(arcsec(2)) equal?

1 Answer

Explanation:

Related questions

Impact of this question

What does $- 2 cos (arctan (6)) + csc (a r c sec (2))$ $-2cos(arctan(6))+csc(arcsec(2))$ equal?