Question

Prove that tan^4A-tan^2A=1, if sin^2A+sin^4A=1

Answer 1

Hey there !

Thanks for the question !

Solution:

$Sin^2 A + Sin^4 A = 1 \\\\ Sin^4 A = 1 - Sin^2 A \\\\ Sin^4A = Cos^2 A => Equation \hspace{1mm} 1 \\\\ \frac{Sin^2 A}{Cos^2A} = \frac{Sin^2A}{Sin^4A} = \frac{1}{Sin^2A} \\\\ => Tan^2A = Cosec^2A \\\\ Converting \hspace{1mm} Cosec^2 A \hspace{1mm} to \hspace{1mm} Cot^2A \hspace{1mm} we \hspace{1mm} get, \\\\ => Tan^2A = 1 + Cot^2A \\\\ Multiplying \hspace{1mm} Tan^2A \hspace{1mm} on \hspace{1mm} both \hspace{1mm} sides \hspace{1mm} we \hspace{1mm} get, \\\\ => Tan^4 A = Tan^2A + ( Tan^2A \times Cot^2A )$

$We \hspace{1mm} know \hspace{1mm} that \hspace{1mm} Tan^2 A \times Cot^2 A = 1 \\\\ So, \hspace{1mm} we \hspace{1mm} get, \\\\ Tan^4A = Tan^2A + 1 \\\\ => \boxed{Tan^4A - Tan^2 A = 1}$

Hence proved !

Hope my answer helped !