Question

Prove under root 1 + cos theta upon 1 minus cos theta is equals to 1 + cos theta upon sin theta

Answer 1

We have to prove that $\sqrt{\dfrac{1+\cos \theta}{1-\cos \theta} } =\dfrac{1+\cos \theta}{\sin \theta} .$

Solution:

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

Rationalising numerator and denominator part, we get

= $\sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}\times \dfrac{1+\cos \theta}{1+\cos \theta} }$

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

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

Using the trigonometric identity:

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

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

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

= R.H.S., proved.

Thus, we have to proved that $\sqrt{\dfrac{1+\cos \theta}{1-\cos \theta} } =\dfrac{1+\cos \theta}{\sin \theta} .$

Answer 2

To prove:

$\bf\sqrt{ \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) } } = \dfrac{1 + \cos( \theta) }{ \sin( \theta) }$

Proof:

LHS:

$\therefore \: \sf \: \sqrt{\dfrac{1 + \cos( \theta) }{1 - \cos( \theta) }}$

Multiplying numerator and denominator with $1 +\cos(\theta)$:

$= \: \sf \: \sqrt{\dfrac{1 + \cos( \theta) }{1 - \cos( \theta) } \times \dfrac{1 + \cos( \theta) }{1 + \cos( \theta) } }$

$= \: \sf \: \sqrt{\dfrac{{ \bigg \{1 + \cos( \theta) \bigg \}}^{2} }{1 - {\cos}^{2} ( \theta) }}$

$= \: \sf \: \sqrt{\dfrac{{ \bigg \{1 + \cos( \theta) \bigg \}}^{2} }{ {\sin}^{2} ( \theta) }}$

$= \: \sf \: \sqrt{ {\bigg \{ \dfrac{ 1 + \cos( \theta) }{ \sin( \theta) } \bigg \}}^{2} }$

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

$\sf = \: RHS$

Hence proved.