Question

if tan a + sin a = m and tan a - sin a = n then show that (m)^2 -(n)^2 = 4√mn​

Answer 1

   

$\Rightarrow\ \boxed{LHS} \\ \\ \Rightarrow\ \boxed{m^2-n^2} \\ \\ \Rightarrow\ \boxed{(\tan A+\sin A)^2-(\tan A-\sin A)^2} \\ \\ \Rightarrow\ \boxed{4\tan A \cdot \sin A} \\ \\ \Rightarrow\ \boxed{4\sqrt{(\tan A \cdot \sin A)^2}} \\ \\ \Rightarrow\ \boxed{4\sqrt{\tan^2A \cdot \sin^2A}} \\ \\ \Rightarrow\ \boxed{4\sqrt{\frac{\sin^2A}{\cos^2A} \cdot \sin^2A}} \\ \\ \\ \Rightarrow\ \boxed{4\sqrt{\frac{1-\cos^2A}{\cos^2A}\cdot \sin^2A}}$

$\Rightarrow\ \boxed{4\sqrt{(\frac{1}{\cos^2A}-1)\sin^2A}} \\ \\ \\ \Rightarrow\ \boxed{4\sqrt{\frac{\sin^2A}{\cos^2A}-\sin^2A}} \\ \\ \\ \Rightarrow\ \boxed{4\sqrt{\tan^2-\sin^2A}} \\ \\ \Rightarrow\ \boxed{4\sqrt{(\tan A+\sin A)(\tan A-\sin A)}} \\ \\ \Rightarrow\ \boxed{4\sqrt{mn}} \\ \\ \Rightarrow\ \boxed{RHS}$

$Hence\ proved\ !!! \\ \\ \\ \boxed{\textit{Thank you...}}$

$\mathfrak{\#adithyasajeevan}$