Question

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

Accepted Answer

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

Answer

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.

Answer

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)

Math

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

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