Mathematics
21.09.2020 17:00
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Select the number line that represents all solutions of x 13.5

Select the number line that represents all solutions of x > 13.5
Select the number line that represents all solutions of x > 13.5
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thedocgalloway
thedocgalloway
4,7(85 marks)

3rd.

Step-by-step explanation:

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brien301
brien301
4,5(3 marks)
A.)

   \csc^2(x) \tan^2 (x)- 1 = \tan^2(x)

Use the identities \csc x = 1 / \sin x and \tan x = \sin x / \cos x on the left-hand side

   \begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

   \begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)}  \\
&=  \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&=  \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}

Use Pythagorean identity for the numerator.

If \sin^2 (x) + \cos^2(x) = 1 then subtracting both sides by \cos^2 (x) yields \sin^2(x) = 1 - \cos^2(x). We can substitute that into the numerator

   \begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}

======

b.)

   \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1

For the left-hand side:
By definition, \sec(x) = 1/\cos(x) and \tan (x) = 1/\cot (x)

   \begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)}  \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}

Since \cot (x) = \cos (x) / \sin (x)

   \begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}

Using Pythagorean identity, \cos^2(x) = 1 - \sin^2(x) so

   \begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}

Prove that:  a.) [csc^2(x)] * [tan^2(x)] -1 = [tan^2 (x)]  b.) sec(x)/cos(x) - tan(x)/cot(x) = 1 sho

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