How to simplify 2sinxcosx+(sinx-cosx)^2/secx ?

Answers

Engr. Ronald

2sin(x)cos(x) + [sin(x) - cos(x)]^2 ----------------------------------------------- ...................sec(x) ...2sin(x)cos(x) + sin^2(x) - 2sin(x)cos(X) + cos^2(x) =--------------------------------------------------------------------- ......................sec(x) ..........1 =-------------- .....sec(x) = cos(x) answer//

la console

= { 2.sin(x).cos(x) + [sin(x) - cos(x)]² } / sec(x) = { 2.sin(x).cos(x) + [sin²(x) - 2.sin(x).cos(x) + cos²(x)] } / sec(x) → recall: sin²(x) + cos²(x) = 1 = { 2.sin(x).cos(x) + [1 - 2.sin(x).cos(x)] } / sec(x) = { 2.sin(x).cos(x) + 1 - 2.sin(x).cos(x) } / sec(x) = 1/sec(x) = cos(x)

az_lender

What you typed means 2*sin(x)*cos(x) + {[sin(x) - cos(x)]^2/sec(x)}, rather than {2*sin(x)*cos(x) + [sin(x)-cos(x)]^2} / sec(x). Anyway, [sin(x) - cos(x)]^2 would be sin^2(x) - 2*sin(x)*cos(x) + cos^2(x) = 1 - 2*sin(x)*cos(x). This suggests that what you MEANT was {2*sin(x)*cos(x) + [sin(x) - cos(x)]^2} / sec(x). So you should learn that when the horizontal bar of a fraction is replaced by a diagonal slash, it is necessary to add symbols of inclusion in order to show the extent of the horizontal bar. Assuming you meant {2*sin(x)cos(x) + [sin(x)-cos(x)]^2} / sec(x), that would be [2*sin(x)*cos(x) + 1 - 2*sin(x)*cos(x)] / sec(x) = 1/sec(x) = cos(x).