Melody

There are lots of answers.

k could be any even number.    ex  2,4,160, etc

k could be any multiple or power of x     ex 6x, 100x, x^2,

k could be any combination of the above.  ex  8x^3ytz

If you have a right angled triangle and all 3 sides you can find either of the missing angles using ANY of the inverse trigonometric ratios.

You can also solve this graphically.

Here is a list of pages that might help you.

Thanks Aziz,

I do these a different way.  My way is perhaps more versatile because you can use it no matter how many terms are inside the brackets.

I take the 1st term in the first bracket and multiply it by the second bracket then

I take the 2nd term in the first bracket and mltiply it by the second bracket then

I can keep doing this until I run out of terms in the first bracket.

Like this

$$\textcolor[rgb]{0,0,1}{(2+4\sqrt{10})}(4\sqrt5 -2)\\\\
=\textcolor[rgb]{0,0,1}{2}(4\sqrt5-2)\textcolor[rgb]{0,0,1}{+4\sqrt{10}}(4\sqrt5-2)\\\\
=8\sqrt5-4+16\sqrt{50}-8\sqrt{10}$$

That is the reciprocal button Aziz. 

I am not sure what you mean by the inverse button.  

The 2nd is in the top left corner like it is with most calculators.

Perhaps you mean inverse trigonometry?

inverse sine is the same as arc sine.    So sin^-1 is the same as asin    

This is probably what you needed to know.

Thank you Rosala,

I always enjoy enjoy your laughs.  

ALSO

Thank you for sharing that special verse with me.  

It is thought provoking you are right.  I definitely like it!

Queen Guinevere 

You give excellent answers Aziz!     

Yours looks a lot shorter than mine Alan  

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

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