Question

Date ID
Page
1- Cose
Clossero - Cost or
at Goso​

Answer 1

Answer:

$\frac{1 - \cos(x) }{1 + \cos(x) } \times \frac{1 - \cos(x) }{1 - \cos(x) } \\ \\ ={ (1 - \cos(x)) {}^{2} \over1 - \cos {}^{2} (x)} \\ \\ = \frac{(1 - \cos(x)) {}^{2} }{ \sin {}^{2} (x) } \\ \\ = \bigg ( \frac{1 - \cos(x) }{ \sin(x) } \bigg) {}^{2} \\ \\ = \bigg ( \frac{1 }{ \sin(x) } - \frac{ \cos(x) }{ \sin(x) } \bigg) {}^{2} \\ \\ { = (\cosec(x) - \cot(x) ) } {}^{2}$

Answer 2

Step-by-step explanation:

Given:

$\frac{1 - \cos( \beta ) }{1 + \cos( \beta ) } = ( \csc( \beta ) - \cot( \beta ) {})^{2}$

By L.H.S:-

$\frac{1 - \cos( \beta ) }{1 + \cos( \beta ) }$

multiplying both numerator and denominator by (1-cos(beta))

$\frac{1 - \cos( \beta ) }{1 + \cos( \beta ) } \times \frac{1 - cos \beta }{1 - \cos( \beta ) }$

$\frac{1 -2 \cos( \beta ) + \cos {}^{2} ( \beta ) }{1 - cos {}^{2} }$

we get;

$\frac{1 + \cos {}^{2} ( \beta ) - 2cos \beta }{ \sin {}^{2} ( \beta ) }$

Since,

$1 - \cos {}^{2} \beta = \sin( \beta )$

$\frac{1}{ \sin {}^{2} ( \beta ) } + \frac{cos {} ^{2} \beta }{ \sin {}^{2} ( \beta ) } - \frac{2 \cos( \beta ) }{ \sin {}^{2} ( \beta ) }$

We know,.

Reciprocal of sin square is cosec square,

cos square ÷sin square is cot square

Therefore,

$\csc {}^{2} ( \beta ) + cot {}^{2} \beta - 2$

Putting

$cot {}^{2} \beta . \ \csc {}^{2} \beta = 1$

$\csc {}^{2} ( \beta ) + \cot^{2} ( \beta ) - 2 \cot( \beta ) \times \csc( \beta )$

but,it is equal to

R.H.S

i.e.

$( \csc {} \beta - cot {}\beta ) {}^{2}$

R.H.S

since,

L.H.S=R.H.S

Hence,proved.