Question

cot theta=2n^2+2mn/m^2+2mn then prove cosec theta=m^2+2mn+2n^2/m^2+2mn

Answer 1

as cot thetha=2n^2+2mn/m^2+2mn then consederind to sides of  2n^2+2mn and m^2+2mn then hypo =root of (2n^2+2mn)+(m^2+2mn)=root of [ 4n^4+8mn^3+4(nm)^2 +m^4+4m^3n+4(nm)^2]

=root of [4n^4+4(nm)^2+m^4+4m^3n+8mn^3+4(nm)^2]
=root of [(m^2+2mn+2n^2)^2]                     as (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca
=m^2+2mn+2n^2=hypo
 therfor cosec thetha =hypo/opposite side=m^2+2mn+2n^2/m^2+2mn

Answer 2

Step-by-step explanation:

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