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2 years ago

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Someone help me please

1 answer:

Ksivusya [100]2 years ago

8 0

Answer:

maybe the answer is 12 units.

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Trig proofs with Pythagorean Identities.

lorasvet [3.4K]

To prove:

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Solution:

$$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$

Multiply first term by $\frac{1+cos x}{1+cos x}$ and second term by $\frac{1-cos x}{1-cos x}$ .

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

Using the identity: $(a-b)(a+b)=(a^2-b^2)$

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

Denominators are same, you can subtract the fractions.

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

Using the identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

$$= \frac{1+\cos^2 x}{\sin^2x}$

Using the identity: $1=\cos ^{2}(x)+\sin ^{2}(x)$

$$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}$

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

$RHS=2 \cot ^{2} x+1$

Using the identity: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

$$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}$

$$=1+2\frac{\cos^{2} x}{\sin^{2} x}$

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

Equation (1) = Equation (2)

LHS = RHS

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Hence proved.

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Step-by-step explanation:

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