Question

hey guys plz help me out with this question​

Answer 1

Answer:

2cosx

Step-by-step explanation:

For a quick solution, consider b=sinx and a=cosx

Thus,

$a.cos2x + b.sin2x +\frac{a^2-b^2}{2b}tan2x\\=a(cos^2x-sin^2x)+b(2.sinx.cosx) + \frac{a^2-b^2}{2b}\frac{2tanx}{1-tan^2x}\\=a(a^2-b^2) + b(2ab)+ \frac{a^2-b^2}{2b}\frac{2ab}{a^2-b^2}\\=a^3-ab^2+2ab^2+a\\=a^3+ab^2+a\\=a(a^2+b^2)+a\\=a(sin^2x+cos^2x)+a\\=2a\\=2cosx$