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Maru [420]
2 years ago
15

Show that sec(x)-tan(x)sin(x)=cos(x)is an identity for any value of x.

Mathematics
1 answer:
Dmitriy789 [7]2 years ago
3 0

Here is one way to show that the left side is identical to the right side.

\sec(x)-\tan(x)\sin(x) = \cos(x)\\\\\\\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}\sin(x) = \cos(x)\\\\\\\frac{1}{\cos(x)}-\frac{\sin^2(x)}{\cos(x)} = \cos(x)\\\\\\\frac{1-\sin^2(x)}{\cos(x)} = \cos(x)\\\\\\\frac{\cos^2(x)}{\cos(x)} = \cos(x)\\\\\\\cos(x)=\cos(x)\\\\\\

Throughout the entire process, the right hand side stayed the same. When working on identities, it's important to keep one side the same while you transform the other side.

Despite the two sides being identical when simplified, there is the issue of cos(x) being zero in the denominator on the left hand side. Be sure to account for this when forming the domain. No such issue occurs on the right hand side. For instance, the value x = pi/2 leads to a division by zero error when in radian mode. So if I was your teacher, I would revise the "for any value of x" and replace it with something along the lines of "for any x value in the domain".

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