Question

Trignometry
Medal would be given
Please Help!!

Answer 1

\[\sec ^{6}x - \tan ^{6}x = 1 + 3\sec ^{2}x \times \tan^{2}x\]

Answer 2

Prove it ^

Answer 3

\(\large \sec ^{6}x - \tan ^{6}x = 1 + 3\sec ^{2}x \times \tan^{2}x\)

\(\large \sec ^{6}x - \tan ^{6}x = 1 + 3(1 + \tan ^{2}x) \times \tan^{2}x\)

\(\large \sec ^{6}x - \tan ^{6}x = 1 + 3(\tan^2 x + \tan ^{4}x)\)

\(\large \sec ^{6}x - \tan ^{6}x = 3 \tan^4 x + 3 \tan^2 x + 1\)

\(\large (1 + \tan^2 x)^3 - \tan ^{6}x = 3 \tan^4 x + 3 \tan^2 x + 1\)

\(\large (\tan^2 x + 1)^3 - \tan ^{6}x = 3 \tan^4 x + 3 \tan^2 x + 1\)

\( \large (\tan^6 x + 3 \tan^4 x + 3 \tan^2 x + 1) - \tan ^{6}x = 3 \tan^4 x + 3 \tan^2 x + 1\)

\( \large 3 \tan^4 x + 3 \tan^2 x + 1 = 3 \tan^4 x + 3 \tan^2 x + 1\)

Answer 4

I appreciate your efforts thanks!

Answer 5

You're welcome.
Here's an explanation.
The first line above is the original problem.
In lines 2-4, I worked on the right side alone.
In line 2, I used the identity \( \sec^2 x = 1 + \tan^2 x\).
In lines 3 and 4, I simplified the right side.
In lines 5-8 I worked only on the left side.
In line 5, I used again the identity \( \sec^2 x = 1 + \tan^2 x\), but I raised it ot the 3rd power, as \( (\sec^2 x)^3 = (1 + \tan^2 x)^3\) or \( \sec^6 x = (1 + \tan^2 x)^3\).
Then in lines 6-8, I worked on the left side to make it look like the right side.
In line 7, I expanded the binomial.
Line 8 was just collecting like terms on the left side.

Answer 6

thanks so much

Answer 7

You're very welcome.