Question #98d02

Question

2138 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-28T22:16:46

The given equality is not an identity.

Take $x = \frac{π}{4}$ $x = pi/4$ :

Left Hand Side $= {sin}^{2} (\frac{π}{4}) + 4 sin (\frac{π}{4}) + \frac{3}{{cos}^{2} (\frac{π}{4})}$ $= sin^2(pi/4)+4sin(pi/4)+3/cos^2(pi/4)$

$= {(\frac{\sqrt{2}}{2})}^{2} + 4 (\frac{\sqrt{2}}{2}) + \frac{3}{{(\frac{\sqrt{2}}{2})}^{2}}$ $= (sqrt(2)/2)^2+4(sqrt(2)/2)+3/(sqrt(2)/2)^2$

$= \frac{1}{2} + 2 \sqrt{2} + 6$ $=1/2 + 2sqrt(2) + 6$

$= \frac{13}{2} + 2 \sqrt{2}$ $= 13/2+2sqrt(2)$

Right Hand Side $= sin (\frac{π}{4}) + \frac{3}{1 - sin (\frac{π}{4})}$ $= sin(pi/4)+3/(1-sin(pi/4))$

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

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

$= \frac{\sqrt{2}}{2} + \frac{12 + 6 \sqrt{2}}{2}$ $=sqrt(2)/2 + (12+6sqrt(2))/2$

$= 6 + \frac{7}{2} \sqrt{2}$ $=6+7/2sqrt(2)$

As the let hand side does not equal the right hand side at $x = \frac{π}{4}$ $x=pi/4$ , the given equality is not an identity.