Question

prove that LHS=RHS.
ans the above question​

Answer 1

Question:-

$: \implies \sqrt{ \dfrac{1 + \cos\theta}{1 - \cos\theta } } = \csc \theta + \cot\theta$

Solution:-

$: \implies \sqrt{ \dfrac{1 + \cos\theta}{1 - \cos\theta } } = \csc \theta + \cot\theta$

Now

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

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

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

Apply this identity in RHS

=> ( a + b ) ²= a² + b² + 2ab

we get

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

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

Using this identities

( a² - b² ) = ( a - b ) ( a + b )

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

$\rm\to \dfrac{1 + \cos \theta }{ \cancel{1 - \cos \theta} } = \dfrac{(1 + \cos {}^{2} \theta + 2 \cos\theta) {}^{} }{ \cancel{( 1 - \cos {}^{} \theta)} (1 + \cos \theta) }$

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

$\rm\to {(1 + \cos \theta)(1 + \cos \theta) }{ } = {(1 + \cos {}^{2} \theta + 2 \cos\theta) {}^{} }$

$\rm\to {(1 + \cos \theta) {}^{2} }{ } = {(1 + \cos {}^{2} \theta + 2 \cos\theta) {}^{} }$

$\rm \to{1 + \cos {}^{2} \theta + 2 \cos\theta {}^{} } = {1 + \cos {}^{2} \theta + 2 \cos\theta {}^{} }$

Hence proved

Some trigonometry identities which we use in this question

$\rm \: \implies \csc \theta = \dfrac{1}{ \sin \theta }$

$\rm \implies \: \cot( \theta) = \dfrac{ \cos( \theta) }{ \sin( \theta) }$

$\rm \: \implies \: 1 - \cos {}^{2} ( \theta) = \sin {}^{2} ( \theta)$

Answer 2

Step-by-step explanation:

$\huge\mathfrak{\color{orange}{\underline{\underline{Answer♡}}}}$

Answer♡

if you will give thanks to this answer then I will give you 1000 free points .I am promising.

refer the attachment

$\huge\underline{ \red{❥L} \green{u}\blue{e} \purple{e} \orange{n}\pink{u}}\:$

$\huge\green{Thanks}$