Question

For what values of x, 0≤x<2π is it true that csc⁡x=√(cot^2⁡x+1)?
a. 0<x<π b. π<x<2π c. π/2≤x≤3π/2
d. 0≤x<π/2, π<x<3π/2 e. None of these

Answer 1

ANSWER on Question #74956 – Math – Trigonometry

For what values of $x$, $0 \leq x < 2\pi$ is it true that

$$\operatorname{csc} x = \sqrt{(\cot^2 x + 1)}?$$ $$a. \quad 0 < x < \pi;$$ $$b. \quad \pi < x < 2\pi;$$ $$c. \quad \frac{\pi}{2} \leq x \leq \frac{3\pi}{2};$$ $$d. \quad 0 \leq x < \frac{\pi}{2}, \quad \pi < x < \frac{3\pi}{2};$$ $$e. \quad \text{None of these}.$$

SOLUTION

Recall some definitions of trigonometry

$$\cot x = \frac{\cos x}{\sin x}$$ $$\operatorname{csc} x = \frac{1}{\sin x}$$ $$\sin x \geq 0 \quad \forall x \in [2\pi n; \pi + 2\pi n], \quad n \in \mathbb{Z}$$

(More information: https://en.wikipedia.org/wiki/List_of_trigonometric_identities)

$$\sin^2 x + \cos^2 x = 1, \quad \forall x \in \mathbb{R}$$

(More information: https://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity)

Just remember the definition of the number modulus

$$|a| = \begin{bmatrix} a, & a \geq 0 \\ -a, & a < 0 \end{bmatrix}$$

(More information: https://en.wikipedia.org/wiki/Absolute_value)

Just remember the properties of the square root

$$\sqrt {x ^ {2}} = | x | = \left[ \begin{array}{l} x, x \geq 0 \\ - x, x < 0 \end{array} \right.$$

(More information: https://en.wikipedia.org/wiki/Square_root)

In our case,

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

Conclusion,

$$\sqrt {\left(\cot^ {2} x + 1\right)} = \frac {1}{| \sin x |}$$

It remains to understand when

$$| \sin x | = \sin x$$

As we know

$$\sin x \geq 0, \forall x \in [ 2 \pi n; \pi + 2 \pi n ], n \in \mathbb {Z}$$

And on the segment $0 \leq x < 2\pi$

$$\sin x \geq 0, 0 \leq x \leq \pi$$

Then,

$$| \sin x | = \sin x, 0 \leq x \leq \pi \rightarrow \boxed {\csc x = \sqrt {\left(\cot^ {2} x + 1\right)}, 0 < x < \pi}$$

Expelled two values $x = 0$ and $x = \pi$ because they $\sin x = 0$ which cannot be, since $\sin x$ is in the denominator of expression

Conclusion,

$$\boxed{\csc x = \sqrt{(\cot^2 x + 1)},\ 0 < x < \pi}$$

ANSWER

$$\boxed{a.\ 0 < x < \pi}$$

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