Prove that 1 + costheta ÷ sintheta - sintheta ÷ 1 + costheta = 2cott

1.	Prove that 1 + costheta ÷ sintheta - sintheta ÷ 1 + costheta = 2cottheta
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, $\sin^2 A+\cos^2 A=1$ ⇒ $\sin^2 A=1-\cos^2 A$ $=\dfrac{1+\cos^2 \theta+2\cos \theta-(1-\cos^2 \theta)}{\sin \theta(1+\cos \theta)}$ $=\dfrac{1+\cos^2 \theta+2\cos \theta-1+\cos^2 \theta}{\sin \theta(1+\cos \theta)}$ $=\dfrac{2\cos^2 \theta+2\cos \theta}{\sin \theta(1+\cos \theta)}$ $=\dfrac{2\cos \theta(\cos \theta+1)}{\sin \theta(1+\cos \theta)}$ $=\dfrac{2\cos \theta}{\sin \theta}$ $= 2\cot \theta$ = R.H.S., proved. Thus, $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta$ , proved.

Prove that 1 + costheta ÷ sintheta - sintheta ÷ 1 + costheta = 2cottheta

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,

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

⇒ $\sin^2 A=1-\cos^2 A$

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

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

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

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

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

$= 2\cot \theta$

= R.H.S., proved.

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