Question

(i) cot 2x + tan x = cosec 2x
​

Answer 1

Answer:

Step-by-step explanation:

Given  cot 2x + tan x = cosec 2x

Consider R.H.S

cot 2x + tan x

$\frac{cos 2x}{sin 2x}$  + $\frac{sinx}{cosx}$                                  {Because cot 2x = $\frac{cos 2x}{sin 2x}$ and tan x = $\frac{sinx}{cosx}$ }

Taking L.C.M  

$\frac{cos2x(cosx)+sin2x(sinx)}{sin2x(cosx)}$--------Equation 1                {I hope u understood how I did}

We know that  

cos 2x = 2 cos²x - 1

sin 2x 2 sin x cos x

Substituting the values in equation 1 we get;

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

$\frac{2cos^{3}x-cosx+2sin^{2}xcosx }{2sinxcos^{2}x }$                    {Removing the brackets and multiplied}  

$\frac{cosx(2cos^{2}x-1+2sin^2x )}{2sinxcos^2x}$                      {Taking cos x common in numerator}

$\frac{2cos^2x-1+2sin^2x}{2sinxcosx}$---Equation2 {Cancelled cos x for numerator and denominator}

We know that

2cos²x - 1 = cos2x----------- Equation 3

cos2x = 1 - 2sin²x

cos2x -1 = -2sin²x

2sin²x = 1-cos2x ----------- Equation 4

Substituting Equation 3 and Equation 4 in Equation 2 we get;

$\frac{cos2x+1-cos2x}{sin2x}$                             {Hope u understood}

$\frac{1}{sin2x}$                                       {cos 2x and -cos 2x will get cancelled}

cosec 2x                                {$\frac{1}{sin2x}$  = cosec 2x}

                                                                                             HENCE PROVED

                                          HOPE YOU UNDERSTOOD                                                                                                          

                                                     THANK YOU