Question

(cosec o - cot o)2 = 1-cos o / 1 + cos o​

Answer 1

Heya!

Here is ur answer..

To proove:

(cosecΦ - cotΦ)² = 1-cosΦ/1+cosΦ

LHS :

= (cosecΦ - cotΦ)²

= (1/sinΦ-cosΦ/sinΦ)²

= (1-cosΦ/sinΦ)²

= (1-cosΦ)²/sinΦ²

= (1-cosΦ)²/1-cos²Φ

= (1-cosΦ)(1-cosΦ) / (1+cosΦ)(1-cosΦ)

= 1-cosΦ/1+cosΦ

RHS = 1-cosΦ/1+cosΦ

Therefore, LHS = RHS

Hence proved!

Identities used:

>> CosecΦ = 1/sinΦ

>> CotΦ = cosΦ/sinΦ

>> sin²Φ = 1-cos²Φ

>> a²-b² = (a+b)(a-b)

>> (a-b)² = (a-b)(a-b)

Hope it helps u..

Answer 2

$\large\sf{\underline{Question:}}$

$\large\sf{Prove\:{( \cosec\:\theta - \cot\:\theta) }^{2} = \frac{1 - \cos \: \theta}{1 + \cos\:\theta } }$

$\large\sf{\underline {Proof:}}$

$\large\sf{\underline{\star{L.H.S.-}}}$

$\large\sf{={( \cosec\:\theta - \cot\:\theta) }^{2}}$

$\large\sf{ ={ ( \frac{1}{ \sin\:\theta } - \frac{ \cos\:\theta }{ \sin} )\: \: }^{2} \:\theta }$

$\large\sf{= \frac{{(1 - \cos \:\theta)}^{2} }{{ \sin } ^{2} \:\theta}}$

$\large\sf{ = \frac{ {(1 - \cos)}^{2} \:\theta }{ { \sin}^{2} \theta } }$

$\large\sf{ = \frac{ {(1 - \cos \:\theta )}^{2} }{ 1 - { \cos }^{2} \theta } }$

$\large\sf{ = \frac{ {(1 - \cos \:\theta )}^{2} }{ {1}^{2} - { \cos }^{2} \theta} }$

$\large\sf{ = \frac{ {(1 - \cos \:\theta )}^{2}}{(1 + \cos\:\theta )(1 - \cos \:\theta )} }$

$\large\sf{=\frac{ (1 - \cos \:\theta )}{(1 + \cos \:\theta)} }$

$\large\sf{Since,L.H.S.=R. H. S. }$

$\large\sf{Hence,\:proved.}$