(tan-sin)square+( 1-cos)square = (1-sec)square? please prove?

Question

(tan-sin)square+( 1-cos)square = (1-sec)square? please prove?

Answer 1

where are the variables? and are you sure the brackets are in the right place?

Answer 2

(tan-sin)square+( 1-cos)square

= tan^2 -- 2sintan + sin^2 + 1 -- 2sec + sec^2

= (tan^2 + 1) + (sin^2 + cos^2) -- 2sec(sin^2 + cos^2)

= sec^2 + 1 -- 2sec

= (1-sec)square

Answer 3

tan-sin = sin/cos   - sin = sin(1/cos -1) = sin/cos (1-cos) = tan (1-cos)

now take (1-cos)^2 common...

(1-cos)^2 [tan^2 +1]

=(1-cos)^2 [sec^2]

=(1-1/sec)^2 [sec^2]

=(sec-1)^2/sec^2  [sec^2]

=(sec-1)^2

=(1-sec)^2