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Sagot :
Answer:
[tex]A.\ \tan(x) \to 2.\ \sin(x) \sec(x)[/tex]
[tex]B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)[/tex]
[tex]C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)[/tex]
[tex]D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)[/tex]
[tex]E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex]
Step-by-step explanation:
Given
[tex]A.\ \tan(x)[/tex]
[tex]B.\ \cos(x)[/tex]
[tex]C.\ \sec(x)csc(x)[/tex]
[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]
[tex]E.\ 2\sec(x)[/tex]
Required
Match the above with the appropriate identity from
[tex]1.\ \sin(x) \tan(x)[/tex]
[tex]2.\ \sin(x) \sec(x)[/tex]
[tex]3.\ \tan(x) + \cot(x)[/tex]
[tex]4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}[/tex]
[tex]5.\ \sec(x) - \sec(x)(\sin(x))^2[/tex]
Solving (A):
[tex]A.\ \tan(x)[/tex]
In trigonometry,
[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]
So, we have:
[tex]\tan(x) = \frac{\sin(x)}{\cos(x)}[/tex]
Split
[tex]\tan(x) = \sin(x) * \frac{1}{\cos(x)}[/tex]
In trigonometry
[tex]\frac{1}{\cos(x)} =sec(x)[/tex]
So, we have:
[tex]\tan(x) = \sin(x) * \sec(x)[/tex]
[tex]\tan(x) = \sin(x) \sec(x)[/tex] --- proved
Solving (b):
[tex]B.\ \cos(x)[/tex]
Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1
So, we have:
[tex]\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}[/tex]
[tex]\cos(x) = \frac{\cos^2(x)}{\cos(x)}[/tex]
In trigonometry:
[tex]\cos^2(x) = 1 - \sin^2(x)[/tex]
So, we have:
[tex]\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}[/tex]
Split
[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}[/tex]
Rewrite as:
[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)[/tex]
Express [tex]\frac{1}{\cos(x)}\ as\ \sec(x)[/tex]
[tex]\cos(x) = \sec(x) - \sec(x) * \sin^2(x)[/tex]
[tex]\cos(x) = \sec(x) - \sec(x)\sin^2(x)[/tex] --- proved
Solving (C):
[tex]C.\ \sec(x)csc(x)[/tex]
In trigonometry
[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]
and
[tex]\csc(x)= \frac{1}{\sin(x)}[/tex]
So, we have:
[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}[/tex]
Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1
[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}[/tex]
[tex]\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}[/tex]
Express [tex]\frac{1}{\cos^2(x)}\ as\ \sec^2(x)[/tex] and [tex]\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}[/tex]
[tex]\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}[/tex]
[tex]\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}[/tex]
In trigonometry:
[tex]tan^2(x) + 1 =\sec^2(x)[/tex]
So, we have:
[tex]\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}[/tex]
Split
[tex]\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}[/tex]
Simplify
[tex]\sec(x)csc(x) = \tan(x) + \cot(x)[/tex] proved
Solving (D)
[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]
Open bracket
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}[/tex]
[tex]1 - \cos^2(x) = \sin^2(x)[/tex]
So, we have:
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}[/tex]
Split
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}[/tex]
[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]
So, we have:
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)[/tex]
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)[/tex] --- proved
Solving (E):
[tex]E.\ 2\sec(x)[/tex]
In trigonometry
[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]
So, we have:
[tex]2\sec(x) = 2 * \frac{1}{\cos(x)}[/tex]
[tex]2\sec(x) = \frac{2}{\cos(x)}[/tex]
Multiply by [tex]\frac{1 - \sin(x)}{1 - \sin(x)}[/tex] --- an equivalent of 1
[tex]2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}[/tex]
[tex]2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}[/tex]
Open bracket
[tex]2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]
Express 2 as 1 + 1
[tex]2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]
Express 1 as [tex]\sin^2(x) + \cos^2(x)[/tex]
[tex]2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]
Rewrite as:
[tex]2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]
Expand
[tex]2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]
Factorize
[tex]2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]
Factor out 1 - sin(x)
[tex]2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]
Express as squares
[tex]2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]
Split
[tex]2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]
Cancel out like factors
[tex]2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex] --- proved
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