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Question

Tanx(sinx + cosx)²+ (1-sec²x)cotx = 2sin²x

Answer 1

I'll work with the left side of the equation:

tan(x)[sin(x) + cos(x)]² + [1 - sec²(x)]cot(x)

Rearranging the identity 1 + tan²(x) = sec²(x) we get 1 - sec²(x) = -tan²(x) and substituting we get:

tan(x)[sin(x) + cos(x)]² + [-tan²(x)]cot(x) then using tan(x)cot(x) = 1 we get:

tan(x)[sin(x) + cos(x)]² - tan(x) then factoring out tan(x) we get:

tan(x)[(sin(x) + cos(x))² - 1] then multiplying out the (sin(x) + cos(x))² we get:

tan(x)[sin²(x) + 2sin(x)cos(x) + cos²(x) - 1] then grouping sin²(x) + cos²(x) and using in²(x) + cos²(x) = 1:

tan(x)[2sin(x)cos(x) + 1 - 1] and simplifying:

2tan(x)sin(x)cos(x) then using tan(x) = sin(x)/cos(x) we get:

2sin²(x)