tg(x) + ctg(x) = 2cosec (2x)

We know that tg(x) = sin(x) /cos(x)

and ctg(x)= cos(x) /sin(x)

Now let us substitute:

sin(x)/cos(x) + cos(x)/sin(x)

Now we need to determine the common denominator:

sin^2 (x) + cos^2(x) / sinxcosx

But we know that sin^2 x+ cos^2 x=1

==> 1/ sinxcosx

Multiply and divide by 2:

==> 2/2sinxcosx

Now 2sinx cosx = sin2x

==> 2/ sin2x

But cosec x= 1/sinx

==> 2/2sin2x = 2 cosec 2x

First, we know that

cosec x = 1/sin x, so that cosec (2x) = 1/sin2x

We'll re-write the equation:

tg(x) + ctg(x) = 2/sin2x

We'll divide by 2 both sides:

[tg(x) + ctg(x)]/2 = 1/sin2x

But ctg x = 1/tg x

So, tg(x) + ctg(x) = tg x + 1/tg x

We'll find the common denominator and we'll add:

tg(x) + ctg(x) = [(tg x)^2 + 1]/tgx

But (tg x)^2 + 1 = 1/(cosx)^2

[(tg x)^2 + 1]/tgx = 1/(tg x)*(cosx)^2

1/(tg x)*(cosx)^2 = cos x/(sin x)*(cos x)^2

After reducing similar terms, we'll get:

cos x/(sin x)*(cos x)^2 = 1/sin x * cos x

[tg(x) + ctg(x)]/2 = 1/2*sin x * cos x = 1/sin 2x

But, from enunciation, [tg(x) + ctg(x)]/2 = 1/sin2x, so the result is verified.

To prove tgx+ctgx = 2cosecx.

Solution:

LHS : tg(x)+ ctg(x) = tg(x)+1/tg(x), as ctg(x) = 1/tg(x)

={(tg(x))^2 + 1}/tag(x)

= (sec(x))^2/tg(x), as (tg(x))^2+1 = (sec(x))^2

= {1/cos(x)}^2/{(sin(x)/cos(x)}

= cos(x)/{(sinx) cos^2(x)}

= 1/sin(x)cos(x)

= 2/{2sin(x) cos(x)}. But 2sin(x )cos(x) = sin(2x)

= 2/sin(2x).

= 2cosecant(2x). As 1/sin(A) = cosec(A)