• Dec 8th 2008, 02:24 PM
SMA777
Verify:

(1+sinx)/cosx + cosx/(1+sinx) = 2secx

Thanks.
• Dec 8th 2008, 05:41 PM
11rdc11
Quote:

Originally Posted by SMA777
Verify:

(1+sinx)/cosx + cosx/(1+sinx) = 2secx

Thanks.

$\displaystyle \frac{1 + \sin{x}}{\cos{x}} + \frac{\cos{x}}{1+\sin{x}}$

$\displaystyle \frac{(1+\sin^2{x}) + \cos^2{x}}{\cos{x}(1 + \sin{x})}$

$\displaystyle \frac{1 + 2\sin{x} +\sin^2{x} + \cos^2{x}}{\cos{x}(1 + \sin{x})}$

$\displaystyle \frac{1 + 2\sin{x} + 1}{\cos{x}(1 + \sin{x})}$

$\displaystyle \frac{2 + 2\sin{x}}{\cos{x}(1 + \sin{x})}$

$\displaystyle \frac{2(1 + \sin{x})}{\cos{x}(1 + \sin{x})}$

$\displaystyle \frac{2}{\cos{x}}$

$\displaystyle 2\sec{x}$
• Dec 8th 2008, 05:41 PM
FrY
$\displaystyle \frac{1+\sin \left( x \right)}{\cos \left( x \right)}+\frac{\cos \left( x \right)}{1+\sin \left( x \right)}=2\sec \left( x \right)$

$\displaystyle \frac{1+2\sin \left( x \right)+\sin ^{2}\left( x \right)+\cos ^{2}\left( x \right)}{\cos \left( x \right)\cdot 1+\sin \left( x \right)}=2\sec \left( x \right)$

$\displaystyle \frac{2+2\sin \left( x \right)}{\cos \left( x \right)\cdot 1+\sin \left( x \right)}=2\sec \left( x \right)$

$\displaystyle \frac{2\cdot \left( 1+\sin \left( x \right) \right)}{\cos \left( x \right)\cdot 1+\sin \left( x \right)}=2\sec \left( x \right)$

$\displaystyle \frac{2}{\cos \left( x \right)}=2\sec \left( x \right)$

$\displaystyle 2\cdot \frac{1}{\cos \left( x \right)}=2\sec \left( x \right)$

$\displaystyle 2\sec \left( x \right)=2\sec \left( x \right)$

(Hi)
• Dec 9th 2008, 10:41 AM
tiar
Quote:

Originally Posted by FrY
$\displaystyle \frac{2+2\sin \left( x \right)}{\cos \left( x \right)\cdot 1+\sin \left( x \right)}=2\sec \left( x \right)$

$\displaystyle \frac{2\cdot \left( 1+\sin \left( x \right) \right)}{\cos \left( x \right)\cdot 1+\sin \left( x \right)}=2\sec \left( x \right)$

(Hi)

Could someone please explain going from $\displaystyle {2+2\sin(x)}$ to $\displaystyle {2\cdot \left( 1+\sin \left( x \right) \right)}$ ?
I am unsure of how this happened.
• Dec 9th 2008, 12:39 PM
FrY
Quote:

Originally Posted by tiar
Could someone please explain going from $\displaystyle {2+2\sin(x)}$ to $\displaystyle {2\cdot \left( 1+\sin \left( x \right) \right)}$ ?
I am unsure of how this happened.

What we did is factorize the expression $\displaystyle 2+2\sin \left( x \right)$ being $\displaystyle 2$ the common factor.