Question

If $x=n\pi -ta{n}^{-1}3$ is a solution of the equation $12tan2x+\frac{\sqrt{10}}{cosx}+1=0$ then

• A. n is any integer
• B. n is an even integer
• C. n is a positive integer
• D. n is an odd integer

Answer 1

Answer: (D)

n is an odd integer

Let n is even
Then for $x=n\pi -ta{n}^{-1}3$

$12tan2x+\frac{\sqrt{10}}{cosx}+1$

$=12tan(2n\pi -2ta{n}^{-1}3)+\sqrt{10}sec(n\pi -ta{n}^{-1}3)+1$

$=-12tan\left(2ta{n}^{-1}3\right)+\sqrt{10}sec\left(ta{n}^{-1}3\right)+1$

$=-12tan(\pi +ta{n}^{-1}\frac{2\times 3}{1-3^2})+\sqrt{10}sec\left(se{c}^{-1}\left(\sqrt{1+3^2}\right)\right)+1$

$=-12tan(\pi +ta{n}^{-1}(-\frac{3}{4}))+\sqrt{10}sec\left(se{c}^{-1}\left(\sqrt{10}\right)\right)+1$

$=-12tanta{n}^{-1}(-\frac{3}{4})+10+1=12\times \frac{3}{4}+10+1=20$

And for odd n

$12tan2x+\frac{\sqrt{10}}{cosx}+1$

$=12tan(2n\pi -2ta{n}^{-1}3)+\sqrt{10}sec(n\pi -ta{n}^{-1}3)+1$

$=-12tanta{n}^{-1}(-\frac{3}{4})-\sqrt{10}sec\left(se{c}^{-1}\left(\sqrt{10}\right)\right)+1=9-10+1=0$

Hence n is odd