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sec^6 (x)- tan^6(x)= 1+3tan^2(x)+3tan^4(x)

Sasini  Jul 26, 2014

Best Answer 

 #3
avatar+91436 
+16

$$\begin{array}{rlll}
[sec^6(x)]-[tan^6(x)]&=&1+[3tan^2(x)]+[3tan^4(x)]\\\\
(\frac{1}{cos^6x})-[(\frac{sin^6x}{cos^6x})]&=&1+[3(\frac{sin^2x}{cos^2x})]+[3(\frac{sin4^x}{cos^4x})]\\\\
cos^6(x)((\frac{1}{cos^6x})-[(\frac{sin^6x}{cos^6x})])&=&cos^6(x)(1+[3(\frac{sin^2x}{cos^2x})]+[3(\frac{sin^4x}{cos^4x})])\\\\
(1)-[sin^6x]&=&cos^6(x)+[3(sin^2xcos^4(x))]+[3(sin^4xcos^2(x))]\\\\
1-[(sin^2x)^3]&=&cos^6(x)+[3(sin^2x)cos^4(x)]+[3((sin^2x)^2)cos^2(x)]\\\\
1-[(1-cos^2x)^3]&=&cos^6(x)+[3(1-cos^2x)cos^4(x)]+[3((1-cos^2x)^2)cos^2(x)]\\\\
1-[(-cos^2x)^3+3(-cos^2x)^2+3(-cos^2x)+1]&=&cos^6(x)+[3(1-cos^2x)cos^4(x)]+[3((cos^2x)^2-2cos^2x+1)cos^2(x)]\\\\
1-[(-cos^6x)+3(cos^4x)-3cos^2x+1]&=&cos^6(x)+[3(cos^4(x)-cos^6(x))]+[3((cos^6x)-2cos^4x+cos^2(x))]\\\\
1+cos^6x-3cos^4x+3cos^2x-1&=&cos^6(x)+[3cos^4(x)-3cos^6(x)]+[3cos^6x-6cos^4x+3cos^2(x)]\\\\
cos^6x-3cos^4x+3cos^2x &=&cos^6(x)+3cos^4(x)-3cos^6(x)+3cos^6x-6cos^4x+3cos^2(x)\\\\
-3cos^4x &=&-3cos^4(x)-6cos^4x\\\\
-3cos^4x &=&-3cos^4x\\\\
LHS&=&RHS\\\\
\end{array}$$

 

$$\mbox{This is true for all values of x}$$

Melody  Jul 26, 2014
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7+0 Answers

 #1
avatar
+3

sec^6(x)-tan^6(x)-3*tan^4(x)-3*tan^2(x)-1=0

Guest Jul 26, 2014
 #2
avatar+91436 
+8

I just did a HUGE page of LaTex on this and lost the whole blink'n lot.

Bummer!!!!!!

ok When I did it on paper I got down to cosx=0

When I did it straight onto the forum I got that it was always true x=anything

I'll redo it a little later.  This is hugely irritating.

Anyway the way I went about it was to change everything to sin and cos

then mult by cos^6x

Then get rid of all the sines

The collect like terms 

and wallah! it will all fall out.

Melody  Jul 26, 2014
 #3
avatar+91436 
+16
Best Answer

$$\begin{array}{rlll}
[sec^6(x)]-[tan^6(x)]&=&1+[3tan^2(x)]+[3tan^4(x)]\\\\
(\frac{1}{cos^6x})-[(\frac{sin^6x}{cos^6x})]&=&1+[3(\frac{sin^2x}{cos^2x})]+[3(\frac{sin4^x}{cos^4x})]\\\\
cos^6(x)((\frac{1}{cos^6x})-[(\frac{sin^6x}{cos^6x})])&=&cos^6(x)(1+[3(\frac{sin^2x}{cos^2x})]+[3(\frac{sin^4x}{cos^4x})])\\\\
(1)-[sin^6x]&=&cos^6(x)+[3(sin^2xcos^4(x))]+[3(sin^4xcos^2(x))]\\\\
1-[(sin^2x)^3]&=&cos^6(x)+[3(sin^2x)cos^4(x)]+[3((sin^2x)^2)cos^2(x)]\\\\
1-[(1-cos^2x)^3]&=&cos^6(x)+[3(1-cos^2x)cos^4(x)]+[3((1-cos^2x)^2)cos^2(x)]\\\\
1-[(-cos^2x)^3+3(-cos^2x)^2+3(-cos^2x)+1]&=&cos^6(x)+[3(1-cos^2x)cos^4(x)]+[3((cos^2x)^2-2cos^2x+1)cos^2(x)]\\\\
1-[(-cos^6x)+3(cos^4x)-3cos^2x+1]&=&cos^6(x)+[3(cos^4(x)-cos^6(x))]+[3((cos^6x)-2cos^4x+cos^2(x))]\\\\
1+cos^6x-3cos^4x+3cos^2x-1&=&cos^6(x)+[3cos^4(x)-3cos^6(x)]+[3cos^6x-6cos^4x+3cos^2(x)]\\\\
cos^6x-3cos^4x+3cos^2x &=&cos^6(x)+3cos^4(x)-3cos^6(x)+3cos^6x-6cos^4x+3cos^2(x)\\\\
-3cos^4x &=&-3cos^4(x)-6cos^4x\\\\
-3cos^4x &=&-3cos^4x\\\\
LHS&=&RHS\\\\
\end{array}$$

 

$$\mbox{This is true for all values of x}$$

Melody  Jul 26, 2014
 #4
avatar+26399 
+13

I think the idea is to prove that the LHS (left-hand side) equals the RHS (right-hand side).

$$$$LHS = \sec^6 x - \tan^6 x$$
$$RHS = 1+3\tan^2 x +3\tan^4 x$$
\\
Divide $\sin^2 x + \cos^2 x = 1$ through by $\cos^2 x$ to get $\tan^2 x + 1 = \sec^2 x$ (remembering that $\sec x= 1/\cos x$)\\
So $\sec^6 x = (\sec^2 x)^3 = (\tan^2 x+1)^3 = \tan^6 x +3\tan^4 x + 3\tan^2 x + 1$\\\\
Therefore $$LHS = \tan^6 x +3\tan^4 x + 3\tan^2 x + 1 - \tan^6 x = 3\tan^4 x + 3\tan^2 x + 1 = RHS$$\\\\
Job done!$$

 

I see Melody beat me to it!

Alan  Jul 26, 2014
 #5
avatar+91436 
+8

Yours looks a lot shorter than mine Alan  

Melody  Jul 26, 2014
 #6
avatar+26399 
+13

No harm in having more than one solution though!

Alan  Jul 26, 2014
 #7
avatar+80931 
+14

Note that sec6x can be written as   [sec2x]3 = [(1 + tan2x)]3

And using the binomial theorem, we have.........

[(1 + tan2x)]3  = [tan2x]3 + 3[tan2x]2 + 3[tan2x] + 1    ...   therefore  ....

sec6x - tan6x =

 [tan2x]3 + 3[tan2x]2 + 3[tan2x] + 1  - tan6x   =

 tan6x  + 3[tan2x]2 + 3[tan2x] + 1 - tan6x   =

3[tan4x] + 3[tan2x] + 1   = which is the same thing as the RHS.....

 

  

CPhill  Jul 26, 2014

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