Prove the following. sec^6x – tan^6x = 1 + 3 sec^2x tan^2x

Question

Prove the following.
$sec^6x – tan^6x = 1 + 3 sec^2x \times tan^2x$

✓

Answer

Taking LHS
$\sec ^6 x-\tan ^6 x$
$=\left(\sec ^2 x\right)^3-\left(\tan ^2 x\right)^3$
$=\left(\sec ^2 x-\tan ^2 x\right)\left(\sec ^4 x+\tan ^2 x \sec ^2 x+\tan ^4 x\right)$
$\left[A s, a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=\sec ^4 x+\tan ^4 x+\tan ^2 x \sec ^2 x+2 \tan ^2 x \sec ^2 x-2 \tan ^2 x \sec ^2 x$
[As, $\sec ^2 \theta-\tan ^2 \theta=1$ ]
$=\sec ^4 x+\tan ^4 x-2 \tan ^2 x \sec ^2 x+3 \tan ^2 x \sec ^2 x$
$=\left(\sec ^2 x-\tan ^2 x\right)^2+3 \tan ^2 x \sec ^2 x\left[a^2+b^2-2 a b=(a-b)^2\right]$
$=1^2+3 \tan ^2 x \sec ^2 x$
$=1+3 \tan ^2 x \sec ^2 x$
$=\text { RHS }$
Proved.

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