Question

I) The set of values of $x$ for which $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$ is $\{n\pi +\frac{\pi }{4},n\in l\}$
II) The expression $(1+tanx+{\tan }^{2}x)$
$(1-\cot x+{\cot }^{2}x)$ is positive for all defined values
of $x$
III)$e^{\sin x}-e^{-s\dot{in}x}\ne 4$ for any real values of $x$.
Which of the following is correct?

• A. only I, II
• B. only II, III
• C. only I, III
• D. I,II,III

Answer 1

The correct option is B only II, III

(1) $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$

or, $\tan (3x-2x)=\tan \frac{\pi }{4}$

or, $\tan x=\tan \frac{\pi }{4}$

or, $x=n\pi +\frac{\pi }{4}$

where $n=0,1,\mathrm{2....}$ i.e. positive integers.

But here given $n\in I$, but it is not true for negative values of $n$.

So, (1) is wrong.

(2)$(1+\tan x+{\tan }^{2}x)(1-\cot x+{\cot }^{2}x)$

$=(\frac{\sin x}{\cos x}+\frac{1}{{\cos }^{2}x})(\frac{1}{{\sin }^{2}x}-\frac{\cos x}{\sin x})$

$=\left(\frac{1+\sin x\cos x}{{\cos }^{2}x}\right)\left(\frac{1-\sin x\cos x}{{\sin }^{2}x}\right)$

$=\frac{1-{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}$

$=\frac{{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}$

This value is always positive.

So, option (2) is correct.

(3)Let's assume

$e^{\sin x}-e^{-\sin x}=4$

$e^{2\sin x}-4e^{\sin x}-1=0$

or, $e^{\sin x}=\frac{4\pm \sqrt{16+4}}{2}=2\pm \sqrt{5}$

or, $\sin x=\ln (2+\sqrt{5})$

and $\sin x=\ln (2-\sqrt{5})$

But $2-\sqrt{5}$ is $-ve$. So, it is not possible.

$\therefore \sin x=\ln (2+\sqrt{5})$

$\Rightarrow 2+\sqrt{5}>e$

$\therefore \ln (2+\sqrt{5})>1$

$\therefore \sin x>1$ which is not possible.

$\therefore e^{\sin x}-e^{-\sin x}\ne 4$

$\therefore$ Option (3) is correct.