Question

(xtana+ycota)(xcota+ytana)=(x+y)^2+4xycot^2a​

Answer 1

Answer:

(x tan a+y cot a)(x cot a+y tan a)=(x+y)^2+4xy cot^2 2a[Proved]

Step-by-step explanation:

LHS=(x tan a+y cot a)(x cot a+y tan a)

=x^2+y^2+xy tan^2 a+xy cot^2 a

=x^2+y^2+xy(tan^2 a+cot^2 a)

=x^2+y^2+xy(4 cot^2 2a+2)

[Since,cot 2a=1/2(cot a-tan a)

or,cot^2 2a=1/4(cot^2 a+tan^2 a-2)(By squaring both sides)

or,4cot^2 2a+2=cot^2 a+tan^2 a]

=(x+y)^2+4xy cot^2 2a

=RHS[Proved]