Question

Prove that
1 + costheta/sintheta-sintheta/1+costheta=2cot theta

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Answer 1

First solve the Lhs by taking the LCM u will get (1+cos theta)^2 -sin theta^2 /(sin theta)x(cos theta)

Then expand the numerator you will get

(1+cos theta)(1+cos theta)-(1-cos^2theta)/(sin theta)(1+cos theta)

Then take 1+cos theta as common in the numerator

(1+cos theta){1+cos theta-(1-cos theta)}/(Sin theta)(1+cos theta)

Then cancel 1+cos theta and open the brackets, you will get

1+cos theta-1+cos theta/sin theta

Then solve the numerator

2cos theta/sin theta

Convert it to cot (cot=cos/sin)

You will get 2cot theta which is equal to the RHS

Answer 2

$\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta$, proved

Step-by-step explanation:

To prove that, $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta$

L.H.S. = $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta}$

Taking LCM of denominator part, we get

$=\dfrac{(1+\cos \theta)^2-\sin^2 \theta}{\sin \theta(1+\cos \theta)}$

Using the algebraic identity,

$(a+b)^{2} =a^{2} +2ab+b^2$

$=\dfrac{1+\cos^2 \theta+2\cos \theta-\sin^2 \theta}{\sin \theta(1+\cos \theta)}$

Using the trigonometric identity,

$\sin^2 A=1-\cos^2 A$

$=\dfrac{1+\cos^2 \theta+2\cos \theta-(1-\cos^2 \theta)}{\sin \theta(1+\cos \theta)}$

$=\dfrac{1+\cos^2 \theta+2\cos \theta-1+\cos^2 \theta}{\sin \theta(1+\cos \theta)}$

$=\dfrac{2\cos^2 \theta+2\cos \theta}{\sin \theta(1+\cos \theta)}$

$=\dfrac{2\cos \theta(\cos \theta+1)}{\sin \theta(1+\cos \theta)}$

$=\dfrac{2\cos \theta}{\sin \theta}$

= $2\cot \theta$

= R.H.S., proved.

Thus, $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta$, proved