sin^2(x)+cos^2(x)+tan^2(x)+cot^2(x)+sec^2(x)+csc^2(x)=7
Note that ...sin^2x + cos^2x = 1 ...so we have
tan^2x + cot^2x + sec^2x + csc^2x = 6
And
1 + tan^2x = sec^2x ....so we have
2tan^2x + cot^2x + csc^2x = 5
And 1 + cot^2x = csc^x so we have
2tan^2x + 2cot^x = 4 divide through by 2
tan^2x + cot^2x = 2
So
sin^2x/ cos^2x + cos^2x/sin^2x = 2
[sin^4x + cos^4] = 2sin^2xcos^2x rearrange
sin^4x - 2sin^2xcos^x + cos^4x = 0 factor
(sin^2x -cos^2x)(sin^2x - cos^2x) =0 setting one of the factors to 0, we have
sin^2x - cos^2x = 0 factor again
(sinx + cosx)(sinx-cosx) = 0
For the first factor, we have sinx = -cosx.... and this occurs at 3pi/4 and at 7pi/4
For the second factor, we have sinx = cos x and this occurs at pi/4 and 5pi/4
So....our solutions are pi/4, 3pi/4, 5pi/4 and 7pi/4
