tanx=b/a then a cos2x+b sin2x =?
- sin2x=2tanx/(tan^2(x) + 1)
- cos2x= (1-tan^2x)/(1 + tan^2x)
Therefore from above formulas :-
sin2x = (2b/a)/( (b/a)^2 + 1) = 2ba/(b^2 + a^2) (after solving the formula you will come to this conclusion)
Similarly cos2x= (1-(b/a)^2)/(1+(b/a)^2) = (a^2-b^2)/(a^2+b^2)
Now subsitute above two conclusions in RHS of given question :-
RHS = a((a^2-b^2))/(a^2+b^2) + b(2ba)/(a^2 + b^2)
{You can see the denominators are same you we can solve the numerators}
RHS = (a^3-ab^2 + 2(b^2)a)/(a^2 + b^2)
Notice the underlined part. They solve to +(b^2)a.
Now RHS = (a^3 + (b^2)a)/(a^2 + b^2)
Therefore Tan x = (a^3 + (b^2)a)/(a^2 + b^2)
Still have doubts ? Feel free to ask in comment section down below.
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