If $$cosx+cos^{2}x=1$$, the numerical value of $$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$ is :
given that : $$cosx+cos^{2}x=1$$
$$cosx+cos^{2}x= sin^2 x +cos^2 x$$
$$sin^2x = cos x $$ .........(1)
Now we need to find value of :
$$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$
Usng equation 1 , $$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$ = $$(cos^{6}x+3cos^{5}x+3sin^{5}x+sin^{6}x-1)$$
= 1 + 3 -1
= 3
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