# Sin^2(x) + tan^2(x) = √2

## Homework Statement

The problem given is sin2(x) + tan2(x) = √2

2. Homework Equations

The relevant equations would be any trigonometric identities

## The Attempt at a Solution

sin2(x) + tan2(x) = √2

sin2(x) + (sin2(x)/cos2(x) ) = √2

[ cos2(x) sin2(x) + sin2(x) ]/ cos2(x) = √2

[ (1- sin2(x)) sin2(x) + sin2(x)] / [ 1 - sin2(x) ] = √2

[ (2 - sin2(x) ) sin2(x) ] / [ 1 - sin2(x) ] = √2

(2 - sin2(x) ) tan2(x) = √2

tan2(x) = ( √2 / [ 2 - sin2(x) ] )

I take this and substitute into the first equation:

sin2(x) + ( √2 / [ 2 - sin2(x) ] ) = √2

( 2 - sin2(x) ) sin2(x) / [ 2 - sin2(x) ] + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 = ( √2 ) 2 - sin2(x)

( 2 - sin2(x) ) sin2(x) + √2 = 2√2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 (1 - sin2(x) )

( ( 2 - sin2(x) ) sin2(x) / (1 - sin2(x) ) )= √2

Here is where I get stuck. I do not know what steps to take next. Please give me hints on this and do not hesitate to point out any mistakes in my work. They are very likely.

fresh_42
Mentor
2021 Award

## Homework Statement

The problem given is sin2(x) + tan2(x) = √2

2. Homework Equations

The relevant equations would be any trigonometric identities

## The Attempt at a Solution

sin2(x) + tan2(x) = √2

sin2(x) + (sin2(x)/cos2(x) ) = √2

[ cos2(x) sin2(x) + sin2(x) ]/ cos2(x) = √2

[ (1- sin2(x)) sin2(x) + sin2(x)] / [ 1 - sin2(x) ] = √2

[ (2 - sin2(x) ) sin2(x) ] / [ 1 - sin2(x) ] = √2
I can follow you until here. What's next is a backward substitution which I don't think will get you very far. At least the next steps are what could be done more easily, because you already have a quadratic equation in ##t := \sin^2(x)## which can be solved.
(2 - sin2(x) ) tan2(x) = √2

tan2(x) = ( √2 / [ 2 - sin2(x) ] )

I take this and substitute into the first equation:

sin2(x) + ( √2 / [ 2 - sin2(x) ] ) = √2

( 2 - sin2(x) ) sin2(x) / [ 2 - sin2(x) ] + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 = ( √2 ) 2 - sin2(x)

( 2 - sin2(x) ) sin2(x) + √2 = 2√2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 (1 - sin2(x) )

( ( 2 - sin2(x) ) sin2(x) / (1 - sin2(x) ) )= √2

Here is where I get stuck. I do not know what steps to take next. Please give me hints on this and do not hesitate to point out any mistakes in my work. They are very likely.

• Charles Link, ForceBoy, SunThief and 1 other person
Wow, I didn't see that! Thank you very much. This was really helpful.