Question

prove that 1+cos/sin-sin/1+cos=2cot​

Answer 1

Answer:(1+cos)^2 - sin^2/sin(1+cos)

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

Cos^2+cos^2+2cos/sin(cos+1) (bcoz 1-sin^2 = cos^2)

2Cos(cos+1)/sin(cos+1)

2cos/sin

2cot

Step-by-step explanation:

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

To take 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,

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

⇒ $\sin^2 A=1-\cos^2 A$

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

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

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

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

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

$= 2\cot \theta$

= R.H.S., proved.

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