Question

Verify the identitiy:

Answer 1

Answer:

$\frac{sinx}{1-cos x}$     =         $cosecx$  +  $cot x$

Step-by-step explanation:

To verify the identity:

sinx/1-cosx = cscx + cotx

we will follow the steps below;

We will take just the left-hand side and work it out to see if it is equal to the right-hand side

sinx/1-cosx

Multiply the numerator and denominator by 1 + cosx

That is;

$\frac{sinx}{1-cos x}$     =    $\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}$

open the parenthesis on the right-hand side of the equation at the numerator and the denominator

sinx(1+cosx) = sinx + sinx cosx

(1-cosx)(1+cosx) = 1 - cos²x

Hence

$\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}$     =  $\frac{sinx + sinx cosx}{1-cos^{2}x }$

But 1- cos²x  = sin²x

Hence we will replace  1- cos²x  by  sin²x

   $\frac{sinx}{1-cos x}$    =       $\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}$     =  $\frac{sinx + sinx cosx}{1-cos^{2}x }$   =  $\frac{sinx+sinxcosx}{sin^{2}x }$

                             

                                  =$\frac{sinx}{sin^{2}x }$   +   $\frac{sinxcosx}{sin^{2}x }$

                                   

                                   =$\frac{1}{sinx}$   +   $\frac{cosx}{sinx}$

             

                                   =$cosecx$  +  $cot x$

$\frac{sinx}{1-cos x}$     =         $cosecx$  +  $cot x$

Note that;

$\frac{1}{sinx}$  = $cosecx$                        

         

 $\frac{cosx}{sinx}$   =       $cot x$