Question : If $\sec x+\cos x=2$, then the value of $\sec^{16} x+\cos^{16} x$ will be:
Option 1: $\sqrt3$
Option 2: $2$
Option 3: $1$
Option 4: $0$
Correct Answer: $2$
Solution : Given: $\sec x+\cos x=2$ We know the algebraic identity, $(a-b)^2=a^2+b^2-2ab$. $\frac{1}{\cos x}+\cos x=2$ Or, $\frac{1+\cos^2 x}{\cos x}=2$ Or, $\cos^2 x–2\cos x+1=0$ Or, $(\cos x-1)^2=0$ $\therefore \cos x=1$ So, $\sec x=1$. The value of $\sec^{16} x+\cos^{16} x$ = $1+1=2$ Hence, the correct answer is $2$.
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