Math  /  Trigonometry

Questiona single trig function with no denominator. 1tan2xsec2x\frac{-1-\tan ^{2} x}{\sec ^{2} x}
Atrempl 1 out of:

Studdy Solution

STEP 1

What is this asking? We're trying to simplify a trigonometric expression, specifically to rewrite 1tan2xsec2x\frac{-1-\tan ^{2} x}{\sec ^{2} x} as a single trig function without a fraction. Watch out! Remember those *trigonometric identities*!
They're our secret weapons here.
Also, don't forget about those pesky negative signs – they can trip you up if you're not careful!

STEP 2

1. Rewrite the numerator
2. Rewrite the denominator
3. Simplify the expression

STEP 3

Let's **factor out** a negative one from the numerator to make things a little clearer.
This gives us: (1+tan2x)sec2x \frac{-(1+\tan^2 x)}{\sec^2 x} Why did we do this?
It helps us see familiar forms and patterns, which is key to using trig identities!

STEP 4

Remember the Pythagorean identity: 1+tan2x=sec2x1+\tan^2 x = \sec^2 x.
So, we can **replace** 1+tan2x1 + \tan^2 x in the numerator with sec2x\sec^2 x: (sec2x)sec2x \frac{-(\sec^2 x)}{\sec^2 x} Now, things are looking much simpler!

STEP 5

We already rewrote the numerator in terms of sec2x\sec^2 x, which matches the denominator.
So, we'll keep the denominator as sec2x\sec^2 x.
Sometimes, the best move is no move!

STEP 6

Now, we have (sec2x)sec2x\frac{-(\sec^2 x)}{\sec^2 x}.
Since the numerator and denominator are the same (except for the negative sign), they **divide to one**, leaving us with: (sec2x)sec2x=1 \frac{-(\sec^2 x)}{\sec^2 x} = -1 Boom! We've simplified our expression to a single trig function (well, technically a constant, which *is* a type of function) without a denominator!

STEP 7

Our simplified expression is 1-1.

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