Question
Prove that:
sec4A (1– sin4A) – 2tan2 A = 1
sec4A (1– sin4A) – 2tan2 A = 1
✓
Answer
Taking LHS
$\begin{array}{l}=\sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A \\
=\sec ^4 A-\sin ^4 A \sec ^4 A-2 \tan ^2 A \\
=\sec ^4 A-\frac{\sin ^4 A}{\cos ^4 A}-2 \tan ^2 A \\
=\sec ^4 A-\tan ^4 A-\tan ^2 A-\tan ^2 A \\
=\sec ^4 A-\tan ^2 A\left(1+\tan ^2 A\right)-\tan ^2 A \\
=\sec ^4 A-\tan ^2 A \sec ^2 A-\tan ^2 A\left[A s, \sec ^2 \theta=1+\tan ^2 \theta\right] \\
=\sec ^2 A\left(\sec ^2 A-\tan ^2 A\right)-\tan ^2 A \\
=\sec ^2 A-\tan ^2 A\left[A s, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right] \\
=1 \\
=\text { RHS }
\end{array}$
Proved!
$\begin{array}{l}=\sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A \\
=\sec ^4 A-\sin ^4 A \sec ^4 A-2 \tan ^2 A \\
=\sec ^4 A-\frac{\sin ^4 A}{\cos ^4 A}-2 \tan ^2 A \\
=\sec ^4 A-\tan ^4 A-\tan ^2 A-\tan ^2 A \\
=\sec ^4 A-\tan ^2 A\left(1+\tan ^2 A\right)-\tan ^2 A \\
=\sec ^4 A-\tan ^2 A \sec ^2 A-\tan ^2 A\left[A s, \sec ^2 \theta=1+\tan ^2 \theta\right] \\
=\sec ^2 A\left(\sec ^2 A-\tan ^2 A\right)-\tan ^2 A \\
=\sec ^2 A-\tan ^2 A\left[A s, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right] \\
=1 \\
=\text { RHS }
\end{array}$
Proved!
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