Question

1 + sin theta- cos theta divided by 1 + sin theta + cos theta whole square = 1 minus cos theta by 1 + cos theta

Answer 1

Here denoting angle Q as A we get,

Taking LHS, [1+sinA−cosA1+sinA+cosA]2Now multiplying and dividing by 1+sinA−cosA, =[1+sinA−cosA1+sinA+cosA×[1+sinA−cosA1+sinA−cosA]]2As we know that (a+b+c)2=a2+b2+c2+2ab+2bc+2ac and for denominator a2−b2=(a−b)(a+b)=[1+sin2A+cos2A+2sinA−2sinAcosA−2cosA1+sin2A+2sinA−cos2A]2=[1+1+2sinA−2sinAcosA−2cosA1−cos2A+sin2A+2sinA]2=[2+2sinA−2sinAcosA−2cosAsin2A+sin2A+2sinA]2=[2(1+sinA)−2cosA(1+sinA)2sinA(sinA+1)]2=[(1−cosA)(1+sinA)sinA(sinA+1)]2=(1−cosA)2sin2A=(1−cosA)21−cos2A=(1−cosA)2(1−cosA)(1+cosA)=(1+cosA)(1+cosA)=RHS