Trigonometry

Given that

$\begin{aligned} \sin x + \cos x & = \sqrt 3 \\ \sqrt 2 \left(\frac 1{\sqrt 2}\sin x + \frac 1{\sqrt 2}\cos x \right) & = \sqrt 3 \\ \sqrt 2 \sin \left(x + \frac \pi 4\right) & = \sqrt 3 \\ \implies \sin \left(x + \frac \pi 4\right) & = \sqrt {\frac 32} > 1 \end{aligned}$

Since $\sin \theta > 1$ has no real solution, there is no real $x$ satisfying the equation above, and hence $\tan x + \cot x$ is not defined for real $x$ .

Let $y = sinx$ . The equation, then, becomes $y + \sqrt{1-y} = \sqrt{3}$ . Squaring both sides twice and simplifying, we get $y^4 - y^2 +1 = 0$ , and this equation doesn’t have any real solutions.

Hi,

The answer is NOT DEFINED because, $-\sqrt{2} \leq sinx + cosx \leq \sqrt{2}$ , For $x \in \Re$

This can be proved by two methods :

• Trignometric

• Using Inequalities

Here I would Prefer using inequalities:

Using Cauchy Schwarz Inequality i.e $(a+b)^2 \leq 2(a^2 + b^2)$

Hence we can say $(sinx + cosx)^2 \leq 2(sin^2 x + cos^2 x)$

As $sin^2 x + cos^2 x = 1$

So $-\sqrt{2} \leq sinx + cosx \leq \sqrt{2}$

Also Your anwer would have come 1 but as $a + \frac{1}{a} \geq 2$

So $tanx + \frac{1}{tanx}$ cannot be 1