Question
Prove that:
$\sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A=1$
$\sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A=1$
✓
Answer
Taking LHS
$=\sec ^4 \mathrm{~A}\left(1-\sin ^4 \mathrm{~A}\right)-2 \tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\sin ^4 \mathrm{~A} \sec ^4 \mathrm{~A}-2 \tan ^2 \mathrm{~A}$
$=\sec ^4 A-\frac{\sin ^4 A}{\cos ^4 A}-2 \tan ^2 A$
$=\sec ^4 \mathrm{~A}-\tan ^4 \mathrm{~A}-\tan ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\tan ^2 \mathrm{~A}\left(1+\tan ^2 \mathrm{~A}\right)-\tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\tan ^2 \mathrm{~A} \sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta\right]$
$=\sec ^2 \mathrm{~A}\left(\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\right)-\tan ^2 \mathrm{~A}$
$=\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=1$
$=\text { RHS }$
Proved!
$=\sec ^4 \mathrm{~A}\left(1-\sin ^4 \mathrm{~A}\right)-2 \tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\sin ^4 \mathrm{~A} \sec ^4 \mathrm{~A}-2 \tan ^2 \mathrm{~A}$
$=\sec ^4 A-\frac{\sin ^4 A}{\cos ^4 A}-2 \tan ^2 A$
$=\sec ^4 \mathrm{~A}-\tan ^4 \mathrm{~A}-\tan ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\tan ^2 \mathrm{~A}\left(1+\tan ^2 \mathrm{~A}\right)-\tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\tan ^2 \mathrm{~A} \sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta\right]$
$=\sec ^2 \mathrm{~A}\left(\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\right)-\tan ^2 \mathrm{~A}$
$=\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=1$
$=\text { RHS }$
Proved!
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