Solve `sin^-1((ax)/c) + sin^-1( (bx)/c) = sin^-1( x )`where a²+b²=c² and c≠0 ?

Solve
`sin^-1((ax)/c) + sin^-1( (bx)/c) = sin^-1( x )`where a²+b²=c² and c≠0

`sin^-1((ax)/c) + sin^-1( (bx)/c) = sin^-1( x )`

`=>sin^-1{((ax)/c)sqrt(1-( (bx)/c)^2)+( (bx)/c)sqrt(1-( (ax)/c)^2)} = sin^-1( x )`

`=>{((ax)/c)sqrt(1-( (bx)/c)^2)+( (bx)/c)sqrt(1-( (ax)/c)^2)}^2 = x^2`

`=>((ax)/c)^2(1-( (bx)/c)^2)+( (bx)/c)^2(1-( (ax)/c)^2)+2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = x^2`

`=>((a^2+b^2))/c^2x^2-(2a^2b^2x^4)/c^4+2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = x^2`

`=>c^2/c^2x^2-(2a^2b^2x^4)/c^4+2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = x^2`

`=>2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = (2a^2b^2x^4)/c^4`

`=>x^2sqrt(1-( (bx)/c)^2) sqrt(1-( (ax)/c)^2) - (abx^4)/c^2=0`

`=>x^2[sqrt(1-( (bx)/c)^2) sqrt(1-( (ax)/c)^2) - (abx^2)/c^2]=0`

so `x^2=0=>x=0`

And

`=>sqrt(1-( (bx)/c)^2) sqrt(1-( (ax)/c)^2)= (abx^2)/c^2`

`=>1-((a^2+b^2))/c^2x^2+cancel((a^2b^2x^4)/c^2)= cancel((a^2b^2x^4)/c^2`

`=>x^2-1=0`

`=>x=+-1`

Ans : `x=0,+1 and -1`

Calling `alpha = (ax)/c` and `beta = (bx)/c` we have

`sin(alpha+beta) = sin alpha cos beta + cos alpha sin beta` so
`sin(alpha+beta)=((ax)/c)sqrt(1-((b x)/c)^2)+((bx)/c)sqrt(1-((ax)/c)^2) = x`

Squaring both sides

`(a^2 x^2)/c^2 + (b^2 x^2)/c^2 - (2 a^2 b^2 x^4)/c^4 + ( 2 a b x^2 sqrt[1 - (a^2 x^2)/c^2] sqrt[1 - (b^2 x^2)/c^2])/c^2=x^2`

Using the fact `a^2+b^2=c^2` we have

`sqrt[c^2 - a^2 x^2] sqrt[c^2 - b^2 x^2]=abx^2`

squaring again

`c^4 - (a^2 + b^2) c^2 x^2=c^4(1-x^2)=0` so finally

`x = pm1` and `x=0` (formerly cancelled).