PLZ help...Trigonometry Identities

Nov 2015
17
0
Baltimore
I have cos(-x)/ sec(x)+tan(-x). So,here what I did.

cos(x)/ (1/cos(x) - (sin(x)/cos(x))
cos(x)/ (1-sin(x)/ cos(x))

cos^2 (x) / (1-sin(x)) <------That's wrong though so where did I go wrong and what's the answer.
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I have cos(-x)/ sec(x)+tan(-x). So,here what I did.

cos(x)/ (1/cos(x) - (sin(x)/cos(x))
cos(x)/ (1-sin(x)/ cos(x))

cos^2 (x) / (1-sin(x)) <------That's wrong though so where did I go wrong and what's the answer.
you did not finish ...

$\dfrac{\cos^2{x}}{1 - \sin{x}}$

$\dfrac{1-\sin^2{x}}{1-\sin{x}}$

$\dfrac{(1-\sin{x})(1+\sin{x})}{1-\sin{x}}$

$1+\sin{x}$
 
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topsquark

Forum Staff
Jan 2006
11,602
3,458
Wellsville, NY
I would add one thing. The initial expression shows that we can't have 1 - sin(x) = 0, since the denominator would "blow up" there. As 1 + sin(x) does not explicitly state this your final answer should.

So I would give the answer as \(\displaystyle 1 + sin(x), ~ sin(x) \neq 1\).

-Dan
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
the finished identity would be ...

$\dfrac{\cos(-x)}{\sec{x}+\tan(-x)} = 1 + \sin{x}$

since the left side is undefined at $x = (2k+1) \cdot \dfrac{\pi}{2} , \, \, k \in \mathbb{Z}$, the value of $\sin{x} \ne \pm 1$
 
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