Question

If $x=n\pi -{tan}^{-1}3$ 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

The correct option is D n is an odd integer

Let n be even

Then $x=n\pi -{\tan }^{-1}3$

Given expression

$12\tan 2x+\frac{\sqrt{10}}{\cos x}+1$

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

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

$(\tan (2n\pi -\theta )=-\tan \theta ,\sec (n\pi -\theta )=\sec \theta ,\text{if n is even})$

$⟹-12\tan (2\pi +{\tan }^{-1}\frac{2\times 3}{1-3^2})+\sqrt{10}\sec \left({\sec }^{-1}\right(\sqrt{1+3^2}\left)\right)+1$

$(2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2})$

$-12\tan (2\pi +{\tan }^{-1}(-\frac{3}{4}))+\sqrt{10}\sec \left({\sec }^{-1}\right(\sqrt{10}\left)\right)+1$

$-12\tan {\tan }^{-1}(-\frac{3}{4})+10+1$

$12\times \frac{3}{4}+10+1=20$

if $x$ is odd

$12\tan 2x+\frac{\sqrt{10}}{\cos x}+1$

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

$(\tan (2n\pi -\theta )=-\tan \theta ,\sec (n\pi -\theta )=-\sec \theta ,\text{if n is odd})$

$-12\tan {\tan }^{-1}(-\frac{3}{4})-\sqrt{10}\sec {\sec }^{-1}\left(\sqrt{10}\right)+1$

$9-10+1=0$

Hence n is Odd