Question

prove each of the following identities:​

Answer 1

hopefully this will help you...

thanks...

Answer 2

Answer:

To keep things tidy, let's write c = cos A and s = sin A.  Then c² + s² = 1.

Also, cot A = cos A / sin A = c/s  and  cosec A = 1/sin A = 1/s.

The first expression is then:

$\displaystyle\frac{\frac{c}s+\frac1s-1}{\frac{c}s-\frac1s+1}=\frac{\left(\frac{c}s+\frac1s-1\right)\times s^2}{\left(\frac{c}s-\frac1s+1\right)\times s^2}=\frac{(c+1-s)s}{(c-1+s)s}$

Now we get the two required expressions by either expanding the numerator or expanding the denominator.

Expanding the numerator gives:

$\displaystyle\frac{(c+1-s)s}{(c-1+s)s}=\frac{(c+1)s-s^2}{(c-1+s)s}\\\\{}\qquad\qquad\quad=\frac{(c+1)s+c^2-1}{(c-1+s)s}\\\\{}\qquad\qquad\quad=\frac{(c+1)s+(c+1)(c-1)}{(c-1+s)s}\\\\{}\qquad\qquad\quad=\frac{(c+1)(s+c-1)}{(c-1+s)s}\\\\{}\qquad\qquad\quad=\frac{c+1}{s}\\\\{}\qquad\qquad\quad=\frac{1+\cos A}{\sin A}$

Expanding the denominator gives:

$\displaystyle\frac{(c+1-s)s}{(c-1+s)s}=\frac{(c+1-s)s}{(c-1)s+s^2}\\\\{}\qquad\qquad\quad=\frac{(c+1-s)s}{(c-1)s+1-c^2}\\\\{}\qquad\qquad\quad=\frac{(c+1-s)s}{-(1-c)s+(1-c)(1+c)}\\\\{}\qquad\qquad\quad=\frac{(c+1-s)s}{(1-c)(-s+1+c)}\\\\{}\qquad\qquad\quad=\frac{s}{1-c}\\\\{}\qquad\qquad\quad=\frac{\sin A}{1-\cos A}$