- If x tan $$60^\circ$$ + cos $$45^\circ$$ = sec $$45^\circ$$ then the value of $$x^{2}$$ + 1 is

Given that x tan $$60^\circ$$ + cos $$45^\circ$$ = sec $$45^\circ$$

We know that value of tan $$60^\circ$$ = $$ \sqrt{3}$$, cos $$45^\circ$$ = $$\frac {1}{\sqrt{2}}$$ and sec $$45^\circ$$ = $$\sqrt{2}$$

=> x$$ \sqrt{3}$$+ $$\frac {1}{\sqrt{2}}$$ = $$\sqrt{2}$$

=> x$$ \sqrt{6}$$ + 1= 2

=> x$$ \sqrt{6}$$ = 1

=> x= $$ \frac{1}{\sqrt{6}}$$

Therefore $$ x^2 + 1 $$ = $$\frac{1} {6} $$ + 1 = $$ \frac{7}{6} $$