Question 1209544
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If sin⁶β + cos⁶β = ¼, 
find (1/sin⁶β) + (1/cos⁶β)
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<pre>
Let  x = {{{sin(beta)}}},  y = {{{cos(beta)}}}.


We have  x^2 + y^2 = 1    (Pythagorean identity)    (1).


Raise (1) to degree 3.  You will get

    x^6 + 3x^4*y^2 + 3x^2*y^4 + y^6 = 1,

    (x^6 + y^6) + 3x^2*y^2(x^2 + y^2) = 1.


Replace here  x^6 + y^6  by 1/4  (since it is given), and replace x^2 + y^2 by 1.  You will get

    {{{1/4}}} + 3x^2*y^2 = 1,

    3x^2*y^2 = 1 - {{{1/4}}} = {{{3/4}}},

    x^2^*y^2 = {{{1/4}}}.


Now   

    {{{1/x^6}}} + {{{1/y^6}}} = {{{(x^6+y^6)/(x^6*y^6)}}} = {{{((1/4))/((1/4)^3)}}} = {{{4^3/4}}} = {{{4^2}}} = 16.


Thus  {{{1/sin^6(beta)}}} + {{{1/cos^6(beta)}}} = 16.


<U>ANSWER</U>.   {{{1/sin^6(beta)}}} + {{{1/cos^6(beta)}}} = 16.
</pre>

At this point, the problem is solved completely.