Prove that1 + costheta/sintheta-sintheta/1+costheta=2cot theta

1.	Prove that1 + costheta/sintheta-sintheta/1+costheta=2cot theta
Answer» $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta$ , 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

Answer»

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