Mathematics, 22.06.2019 12:50 baller560

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation 4. 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 from: NEUROPHARMACOLOGICAL

(sin x)/(1-cosx) + sinx/(1-cosx) =

2 csc x

We have like denominators.

Just add the numerators.

(2 sin x)/(1 - cos x) = 2 csc x

[1/(1 - cos x)](2 sin x) = 2 csc x

Take it from here.

Answer from: Leahma327

a ) sin x ( \frac{sinx}{cosx}*cosx- \frac{cosx}{sinx}+*cos x)= \\ sinx(sinx- \frac{cos^{2} x}{sin^{2} x})=
sin^{2}x-cos^{2} x=1-cos^{2}x-cos^{2} x= \\ 1-2cos^{2}x

b ) 1 + \frac{1}{cos^{2} x} *sin^{2} x=1+ \frac{sin^{2} x}{cos^{2} x} = \\ \frac{cos {2} x+sin x^{2} x}{cos^{2} x} = \frac{1}{cos^{2} x} =sec^{2} x

c) \frac{sinx}{1-cosx} + \frac{sinx}{1+cosx}= \frac{sin x ( 1+cosx)+sinx(1-cosx)}{1-cos^{2} x} = \\ \frac{sinx+sinxcosx+sinx-sinxcosxx}{1-cos^{2}x }= \\ \frac{2sinx}{sin^{2} x} = \frac{2}{sinx}=2 csc x

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

Step-by-step explanation:

Answer from: clonetrooper099

(sin(x)/(1-cosx))+sinx/(1+cosx)=2cscx

multiply both parts the left hand side by (1+cosx) and (1-cosx) respectively to get:

(sinx(1+cosx))+sinx(1-cosx))/(1-cos^2(x))

Simplify by the pythagorean identity, factoring and basic algebra:

(sin(x)(1+cosx+1-cosx))/(sin^2(x))

2/sin(x)=2cscx

Answer from: 375025

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

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

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

Answer from: BraedynM7866

Now, recall that the conjugate of a + b, is just a - b, therefore, the conjugate of 1 - sin(x) is just 1 + sin(x).

so hmmm let's multiply the left-side by the conjugate of the numerator then.

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\$

$\bf \cfrac{1-sin(x)}{cos(x)}=\cfrac{cos(x)}{1+sin(x)}
\\\\\\
\cfrac{1-sin(x)}{cos(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \cfrac{[1-sin(x)][1+sin(x)]}{cos(x)[1+sin(x)]}
\\\\\\
\cfrac{1^2-sin^2(x)}{cos(x)[1+sin(x)]}\implies \cfrac{\underline{cos^2(x)}}{\underline{cos(x)}[1+sin(x)]}\implies \cfrac{cos(x)}{1+sin(x)}$

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