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avanturin [10]
4 years ago
15

Verify that this trigonometric equation is an identity?

Mathematics
1 answer:
Vlada [557]4 years ago
8 0
\cot x\sec^4 x=\cot x+2\tan x+\tan^3x\\\\L=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos^4x}=\dfrac{1}{\sin x}\cdot\dfrac{1}{\cos^3x}=\dfrac{1}{\sin x\cos^3x}\\\\R=\dfrac{\cos x}{\sin x}+2\cdot\dfrac{\sin x}{\cos x}+\dfrac{\sin^3x}{\cos^3x}\\\\=\dfrac{\cos x\cos^3x}{\sin x\cos^3x}+\dfrac{2\sin x\cos^2x}{\cos x\sin x\cos^2x}+\dfrac{\sin^3x\sin x}{\cos^3x\sin x}\\\\=\dfrac{\cos^4x+2\sin^2x\cos^2x+\sin^4x}{\sin x\cos^3x}\\\\=\dfrac{(\cos^2x)^2+2\sin^2x\cos^2x+(\sin^2x)^2}{\sin x\cos^3x}

=\dfrac{(\cos^2x+\sin^2x)^2}{\sin x\cos^3x}=\dfrac{1}{\sin x\cos^3x}=L\\\\Used:\\\tan(a)=\dfrac{\sin(a)}{\cos(a)}\\\cot(a)=\dfrac{\cos(a)}{\sin(a)}\\\sec(a)=\dfrac{1}{\cos(a)}\\\sin^2a+\cos^2a=1\\(a+b)^2=a^2+2ab+b^2
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Answer:

  see below

Step-by-step explanation:

Put -1 where x is in each expression and evaluate it.

__

You will find that the expression is zero when the numerator is zero. And you will find the numerator is zero when it has a factor that is equivalent to ...

  (x +1)

Substituting x=-1 into this factor makes it be ...

  (-1 +1) = 0

__

Evaluating the first expression, we have ...

\dfrac{4(x+1)}{(4x+5)}=\dfrac{4(-1+1)}{4(-1)+5}=\dfrac{4\cdot 0}{1}=0

This first expression is one you want to "check."

You can see that the reason the expression is zero is that x+1 has a sum of zero. You can look for that same sum in the other expressions. (The tricky one is the one with the factor (x -(-1)). You know, of course, that -(-1) = +1.)

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