Question

If $x_1$ and $x_2$ are two real solutions of the equation $(x{)}^{\ln x^2}=e^{18},$ then the product $(x_1.x_2)$ equals

• A. $\frac{\left({\cot }^2{5}^{\circ }\right)\left({\cos }^2{5}^{\circ }\right)}{{\cot }^2{5}^{\circ }-{\cos }^2{5}^{\circ }}$
• B. $\frac{\left({\tan }^2{10}^{\circ }\right)\left({\sin }^2{10}^{\circ }\right)}{{\tan }^2{10}^{\circ }-{\sin }^2{10}^{\circ }}$
• C. $\sec 0+\sec \frac{\pi }{7}+\sec \frac{2\pi }{7}+\sec \frac{3\pi }{7}+\sec \frac{4\pi }{7}+\sec \frac{5\pi }{7}+\sec \frac{6\pi }{7}=1$
• D. $\sqrt{1-{\sin }^2{110}^{\circ }}\cdot \sec {110}^{\circ }$

Answer 1

Answer: (A), (B), (C)

$\sec 0+\sec \frac{\pi }{7}+\sec \frac{2\pi }{7}+\sec \frac{3\pi }{7}+\sec \frac{4\pi }{7}+\sec \frac{5\pi }{7}+\sec \frac{6\pi }{7}=1$

$(x{)}^{\ln x^2}=e^{18}$
$\Rightarrow \ln x^2\times \ln x=18$
$\Rightarrow 2(\ln x{)}^2=18$
$\Rightarrow (\ln x{)}^2=9\Rightarrow (\ln x)=\pm 3$
$\Rightarrow x_1=e^3,x_2=e^{-3}$
$\therefore x_1.x_2=1$

Now,
$\frac{\left({\cot }^2{5}^{\circ }\right)\left({\cos }^2{5}^{\circ }\right)}{{\cot }^2{5}^{\circ }-{\cos }^2{5}^{\circ }}=\frac{{\cos }^4{5}^{\circ }}{{\cos }^2{5}^{\circ }(1-{\sin }^2{5}^{\circ })}=1$

$\frac{\left({\tan }^2{10}^{\circ }\right)\left({\sin }^2{10}^{\circ }\right)}{{\tan }^2{10}^{\circ }-{\sin }^2{10}^{\circ }}=\frac{{\sin }^4{10}^{\circ }}{{\sin }^2{10}^{\circ }(1-{\cos }^2{10}^{\circ })}=1$

$\sec 0+\sec \frac{\pi }{7}+\sec \frac{2\pi }{7}+\sec \frac{3\pi }{7}+\sec \frac{4\pi }{7}+\sec \frac{5\pi }{7}+\sec \frac{6\pi }{7}$
$=1-\sec \frac{6\pi }{7}-\sec \frac{5\pi }{7}-\sec \frac{4\pi }{7}+\sec \frac{4\pi }{7}+\sec \frac{5\pi }{7}+\sec \frac{6\pi }{7}=1$

$\sqrt{1-{\sin }^2{110}^{\circ }}\cdot \sec {110}^{\circ }$
$=|\cos {110}^{\circ }|\times \sec {110}^{\circ }=-\cos {110}^{\circ }\times \sec {110}^{\circ }=-1$