if 2tan^2a tan^2b tan^2c+tan^2atan^2b+tan^2btan^2c+tan^2a+tan^c=1 prove that sin^2a + sin^2b + sin^2c=1
2(tan^{2} a)(tan^{2} b)(tan^{2} c) + (tan^{2} a)(cot^{2} b) + (tan^{2} b)(tan^{2} c) + (tan^{2} a) + tan^{c} = 1
Just thought I should make sure the last term, tan c, is correctly written since everything else was squared, and didn't involve the unknown as the exponent. I'm guessing you meant tan^{2} c.
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