જો sin x + {sin ^2}x = 1, તો {cos ^{12} | vedclass

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(c) We have, $\sin x + {\sin ^2}x = 1$

or $\sin x = 1 - {\sin ^2}x$ or $\sin x = {\cos ^2}x$  

$	herefore$ ${\cos ^{12}}x + 3{\cos ^{

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(c) We have, $\sin x + {\sin ^2}x = 1$

or $\sin x = 1 - {\sin ^2}x$ or $\sin x = {\cos ^2}x$  

$	herefore$ ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 2$

$ = {\sin ^6}x + 3{\sin ^5}x + 3{\sin ^4}x + {\sin ^3}x - 2$

$ = {({\sin ^2}x)^3} + 3{({\sin ^2}x)^2}\sin x + 3({\sin ^2}x){(\sin x)^2} + {(\sin x)^3} - 2$ 

$ = {({\sin ^2}x + \sin x)^3} - 2$

$ = {(1)^3} - 2$          $[ \because \sin x + {\sin ^2}x = 1({m{given}})]$

$= -1.$

જો $\sin x + {\sin ^2}x = 1$, તો ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 2 =$

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