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eduard
3 years ago
13

Solve the inequality e-2 <0

Mathematics
1 answer:
Lena [83]3 years ago
8 0

Answer:

e < 2

Step-by-step explanation:

Given

e - 2 < 0 ( ad 2 to both sides )

e < 2

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Step-by-step explanation:

20 x 2 =40

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Sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x
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Answer:

The answer in the procedure

Step-by-step explanation:

we have

\frac{sin(x)}{1-cos(x)}+\frac{sin(x)}{1-cos(x)}=2csc(x)

Prove the identity

In this problem there is a mistake: in order to obtain an identity the second denominator at the left side must be 1 + cosx

so

\frac{sin(x)}{1-cos(x)}+\frac{sin(x)}{1+cos(x)}=2csc(x)

Adds fraction in the left side

\frac{sin(x)(1+cos(x)+sin(x)(1-cos(x)}{1-cos^{2} (x)}=2csc(x)\\ \\\frac{2sin(x)}{1-cos^{2} (x)}=2csc(x)

Remember that

csc(x)=\frac{1}{sin(x)}

and

sin^{2}(x)+cos^{2}(x)=1

sin^{2}(x)=1-cos^{2}(x)

substitute

\frac{2sin(x)}{sin^{2}(x)}=2\frac{1}{sin(x)}

Multiply both sides by sin(x)

(sin(x))\frac{2sin(x)}{sin^{2}(x)}=2

\frac{2sin^{2}(x)}{sin^{2}(x)}=2

2=2 ----> identity verified

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