Question

solve correctly.......​

Answer 1

Given :-

An equation, $\bf \sqrt{ \dfrac{1 + cos \: \theta}{1 - cos \: \theta}} + \sqrt{ \dfrac{1 - cos \: \theta}{1 + cos \: \theta}} = 2 \: cosec \: \theta$

To Prove :-

$\bf \sqrt{ \dfrac{1 + cos \: \theta}{1 - cos \: \theta}} + \sqrt{ \dfrac{1 - cos \: \theta}{1 + cos \: \theta}} = 2 \: cosec \: \theta$

Proof :-

Taking L.H.S.,

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

Now, rationalize the denominator,

$\implies \bf \bf \sqrt{ \dfrac{1 + cos \: \theta}{1 - cos \: \theta} \times \dfrac{1 + cos \: \theta}{1 + cos \: \theta} } + \sqrt{ \dfrac{1 - cos \: \theta}{1 + cos \: \theta} \times \dfrac{1 - cos \: \theta}{1 - cos \: \theta} }$

Using identity in the denominator,

$\boxed{ \bf{ \pink{(a + b)(a - b) = {a}^{2} - {b}^{2} }}}$

$\implies \bf \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ {(1)}^{2} - {(cos \: \theta)}^{2} } } + \sqrt{ \dfrac{ {(1 - cos \: \theta)}^{2} }{ {(1)}^{2} - {(cos \: \theta)}^{2} } } \\ \\ \\ \implies \bf \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ (1 - {cos \:}^{2} \theta )} } + \sqrt{ \dfrac{ {(1 - cos \: \theta)}^{2} }{ (1 - {cos}^{2} \: \theta)} } \\ \\ \\ \implies \bf \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ {sin}^{2} \theta } } + \sqrt{ \dfrac{ {(1 - cos \: \theta)}^{2} }{ {sin}^{2} \: \theta} } \\ \\ \\ \implies \bf \dfrac{1 + cos \: \theta}{sin \: \theta} + \dfrac{1 - cos \: \theta}{sin \: \theta} \\ \\ \\ \implies \bf \dfrac{(1 + cos \: \theta) + (1 - cos \: \theta)}{sin \: \theta} \\ \\ \\ \implies \bf \dfrac{1 + 1}{sin \: \theta} \\ \\ \\ \implies \bf \dfrac{2}{sin \: \theta}$

We know that,

$\: \: \bf \bullet \: \: \: cosec \: \theta = \dfrac{1}{sin \: \theta}$

Therefore,

$\implies \bf 2 \: cosec \: \theta \\ \\ \\ \bf \implies R.H.S.$

Hence Proved !

Note:-

• learn all trigonometric identity for better understanding.

Answer 2

Answer:

Proof :-

Taking L.H.S.,

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

Now, rationalize the denominator,

$\implies \bf \bf \sqrt{ \dfrac{1 + cos \: \theta}{1 - cos \: \theta} \times \dfrac{1 + cos \: \theta}{1 + cos \: \theta} } + \sqrt{ \dfrac{1 - cos \: \theta}{1 + cos \: \theta} \times \dfrac{1 - cos \: \theta}{1 - cos \: \theta} }$

Using identity in the denominator,

$\boxed{ \bf{ \orange{(a + b)(a - b) = {a}^{2} - {b}^{2} }}}$

$\implies \bf \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ {(1)}^{2} - {(cos \: \theta)}^{2} } } + \sqrt{ \dfrac{ {(1 - cos \: \theta)}^{2} }{ {(1)}^{2} - {(cos \: \theta)}^{2} } } \\ \\ \\ \implies \bf \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ (1 - {cos \:}^{2} \theta )} } + \sqrt{ \dfrac{ {(1 - cos \: \theta)}^{2} }{ (1 - {cos}^{2} \: \theta)} } \\ \\ \\ \implies \bf \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ {sin}^{2} \theta } } + \sqrt{ \dfrac{ {(1 - cos \: \theta)}^{2} }{ {sin}^{2} \: \theta} } \\ \\ \\ \implies \bf \dfrac{1 + cos \: \theta}{sin \: \theta} + \dfrac{1 - cos \: \theta}{sin \: \theta} \\ \\ \\ \implies \bf \dfrac{(1 + cos \: \theta) + (1 - cos \: \theta)}{sin \: \theta} \\ \\ \\ \implies \bf \dfrac{1 + 1}{sin \: \theta} \\ \\ \\ \implies \bf \dfrac{2}{sin \: \theta}$

We know that,

$\: \: \bf \bullet \: \: \: cosec \: \theta = \dfrac{1}{sin \: \theta}$

Therefore,

$\implies \bf 2 \: cosec \: \theta \\ \\ \\ \bf \implies R.H.S.$

Hence Proved !