Question

prove that (1+sin theta+cos theta)^2=2(1+sin theta)(1+cos theta)

Answer 1

Your answer is ---

LHS = (1+sin∅+cos∅)^2

= {1+(sin∅+cos∅)}^2

= 1 + (sin∅+cos∅)^2 + 2(sin∅+cos∅)

= 1 + sin^2∅+cos^2∅+2sin∅cos∅+2(sin∅+cos)

= 1+1 + 2sin∅cos∅+2(sin∅+cos)

= 2 + 2sin∅cos∅+2(sin∅+cos)

= 2(1+sin∅cos∅+sin∅cos∅)

= 2(1+sin∅)(1+cos∅) [°•° 1+sin∅cos∅+sin∅cos∅ = (1+sin∅)(1+cos∅)]

= 2(1+sin∅)(1+cos∅) = RHS

Hence proved