Math  /  Trigonometry

Question7. Simplify each expression. a) 2(csc2xcot2x)2\left(\csc ^{2} x-\cot ^{2} x\right) b) cot2x(sec2x1)\cot ^{2} x\left(\sec ^{2} x-1\right) d) cosxsinxcotx\frac{\cos x}{\sin x \cot x}

Studdy Solution

STEP 1

What is this asking? We need to simplify three trigonometric expressions using trig identities. Watch out! Remember those Pythagorean identities, they're crucial here!
Also, don't forget the basic definitions of your trig functions.

STEP 2

1. Simplify 2(csc2xcot2x)2(\csc^2 x - \cot^2 x)
2. Simplify cot2x(sec2x1)\cot^2 x (\sec^2 x - 1)
3. Simplify cosxsinxcotx\frac{\cos x}{\sin x \cdot \cot x}

STEP 3

Alright, let's tackle this first expression!
We've got 2(csc2xcot2x)2(\csc^2 x - \cot^2 x).
Do you see something familiar?
It involves csc2x\csc^2 x and cot2x\cot^2 x, which reminds us of the **Pythagorean identity**: 1+cot2x=csc2x1 + \cot^2 x = \csc^2 x.

STEP 4

Let's **rewrite** the identity to match what we have in the expression: If we subtract cot2x\cot^2 x from both sides of 1+cot2x=csc2x1 + \cot^2 x = \csc^2 x, we get 1=csc2xcot2x1 = \csc^2 x - \cot^2 x.
Perfect!

STEP 5

Now, **substitute** this back into our original expression: 2(csc2xcot2x)2(\csc^2 x - \cot^2 x) becomes 212 \cdot 1, which simplifies to **2**.
Boom!

STEP 6

On to the next one: cot2x(sec2x1)\cot^2 x (\sec^2 x - 1).
This time, we see sec2x\sec^2 x, which screams another **Pythagorean identity**: 1+tan2x=sec2x1 + \tan^2 x = \sec^2 x.

STEP 7

Let's **rearrange** this identity to get sec2x1=tan2x\sec^2 x - 1 = \tan^2 x.
Now, **substitute** this back into our expression: cot2x(sec2x1)\cot^2 x (\sec^2 x - 1) becomes cot2xtan2x\cot^2 x \cdot \tan^2 x.

STEP 8

Remember that cotx=1tanx\cot x = \frac{1}{\tan x}, so cot2x=1tan2x\cot^2 x = \frac{1}{\tan^2 x}.
So, our expression becomes 1tan2xtan2x\frac{1}{\tan^2 x} \cdot \tan^2 x.

STEP 9

Multiplying these together, we get tan2xtan2x\frac{\tan^2 x}{\tan^2 x}, which simplifies to **1**.
Fantastic!

STEP 10

Last but not least, we have cosxsinxcotx\frac{\cos x}{\sin x \cdot \cot x}.
We know that cotx=cosxsinx\cot x = \frac{\cos x}{\sin x}.
Let's **substitute** that in: cosxsinxcosxsinx\frac{\cos x}{\sin x \cdot \frac{\cos x}{\sin x}}.

STEP 11

This simplifies to cosxsinxcosxsinx\frac{\cos x}{\frac{\sin x \cdot \cos x}{\sin x}}.
Now, we can multiply the numerator and denominator by sinx\sin x to get cosxsinxsinxcosx\frac{\cos x \cdot \sin x}{\sin x \cdot \cos x}.

STEP 12

This simplifies to sinxcosxsinxcosx\frac{\sin x \cdot \cos x}{\sin x \cdot \cos x}.
Anything divided by itself is 1, so our final answer is **1**.
We did it!

STEP 13

a) 2(csc2xcot2x)=22(\csc^2 x - \cot^2 x) = 2 b) cot2x(sec2x1)=1\cot^2 x (\sec^2 x - 1) = 1 d) cosxsinxcotx=1\frac{\cos x}{\sin x \cdot \cot x} = 1

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