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# plz.. solve this questions

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

Sasini  Jul 26, 2014

#3
+91972
+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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#1
+3

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

Guest Jul 26, 2014
#2
+91972
+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
+91972
+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
#4
+26550
+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 xRHS = 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
+91972
+8

Yours looks a lot shorter than mine Alan

Melody  Jul 26, 2014
#6
+26550
+13

No harm in having more than one solution though!

Alan  Jul 26, 2014
#7
+84385
+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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