Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

# Does anyone know how to solve [1+tan(2)x / tan(2)x]=csc(2)x?

### 3 Answers

Relevance
• Favorite Answer

The equation is not quite clear. I suppose that you mean this:

[1 + tan(2x)] / [tan(2x)] = csc(2x)

If yes, then

cotg(2x) + 1 = csc(2x)

cos(2x) + sin(2x) = 1

(√2) sin(2x + π/4) = 1

sin(2x + π/4) = √2/2

2x + π/4 = arcsin(√2/2)

Solutions are

x = πn

and

x = π/4 + πn

where n is any whole number.

-

Note: To square both sides of an equation is not a correct way. It gives additional false results.

-

• If the (2) was to mean squared then your statement is an identity, cot^2( x )+ 1 =

csc^2 (x). If it meant 2x then it can be written as cos x + sin x = 1. Square both sides ,clean it up, you find sin(4x) = 0 which has answers n(pi / 4) , n = 0,1,2,3,4,5,6,7 for answers in [0,2 pi)

• [ 1 + tan(2)x / tan(2)x ] = csc(2)x

we have : tan2x = sin2x / cos2x

we have

1 +tan2x =( csc2x)(tan2x) = (1 /sin2x)(sin2x / (cos2x)

1 +tan2x = 1 / cos2x

cos2x( 1 +tan2x ) = 1

cos2x + (tan2x) (cos2x) = 1

cos2x +[( sin2x / cos2x)(cos2x)] = 1

cos2x + sin2x = 1

we have ( csc2x = ( 1 / sin2x) =( csc2x) ( sin2x) = 1

we have

cos2x + sin2x =( csc2x) (sin2x)

we have

csc2x = [( cos2x + sin2x ) / sin2x ]

= (cos2x / sin2x) + (sin2x /sin2x)

but ( cot (2x )= ( cos2x / sin2x)

csc2x = cot 2x + 1

but ; cot 2x = 1 / tan2x

csc2x = ( 1 / tan2x) + 1

csc2x = (1 + tan2x) / tan2x =

(1 + tan2x ) / tan2x = csc2x

Still have questions? Get your answers by asking now.