Question

2tan2x=cosx+sinx /cosx-sinx - cosx-sinx/cosx+sinx

Answer 1

Multiply & divide the first fraction by (cosx + sinx) and the second by (cosx - sinx). 

Therefore: 

(cosx + sinx)/(cosx - sinx) - (cosx - sinx)/(cosx + sinx) 

= (cosx + sinx)^2/[cos^2(x) - sin^2(x)] - 

(cosx - sinx)^2/[cos^2(x) - sin^2(x)]. 

Now bring it to the common denominator [cos^2(x) - sin^2(x)] = cos2(x) and expand the square terms in the numerator, simplify 

[cos^2(x) + 2cosx*sinx + sin^2(x) - cos^2(x) + 2cosx*sinx - sin^2(x)]/cos(2x) 

= 2sin(2x)/cos(2x) 

= 2tan(2x)
taking l.c.m, we get : 
[(cos x + sin x)^2 - (cos x - sin x)^2] / [(cos x)^2 - (sin x)^2] 
numerator, by a^2 - b^2 = (a+b)x(a-b) becomes (2cos x)(2sin x) 
denominator becomes cos 2x by known trig reduction 
further, by 2sin x cos x = sin 2x, 
we get 2sin 2x/cos 2x = 2 tan 2x

Answer 2

Step-by-step explanation:

Multiply & divide the first fraction by (cosx + sinx) and the second by (cosx - sinx).  

Therefore:  

(cosx + sinx)/(cosx - sinx) - (cosx - sinx)/(cosx + sinx)  

= (cosx + sinx)^2/[cos^2(x) - sin^2(x)] -  

(cosx - sinx)^2/[cos^2(x) - sin^2(x)].  

Now bring it to the common denominator [cos^2(x) - sin^2(x)] = cos2(x) and expand the square terms in the numerator, simplify  

[cos^2(x) + 2cosx*sinx + sin^2(x) - cos^2(x) + 2cosx*sinx - sin^2(x)]/cos(2x)  

= 2sin(2x)/cos(2x)  

= 2tan(2x)

taking l.c.m, we get :  

[(cos x + sin x)^2 - (cos x - sin x)^2] / [(cos x)^2 - (sin x)^2]  

numerator, by a^2 - b^2 = (a+b)x(a-b) becomes (2cos x)(2sin x)  

denominator becomes cos 2x by known trig reduction  

further, by 2sin x cos x = sin 2x,  

we get 2sin 2x/cos 2x = 2 tan 2x