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Math  /  Trigonometry

Question30. cos⁡x1−sin⁡x=sec⁡x+tan⁡x\frac{\cos x}{1-\sin x}=\sec x+\tan x

Studdy Solution

STEP 1

What is this asking? We need to prove that dividing cosine of xx by one minus sine of xx equals the sum of secant of xx and tangent of xx. Watch out! Remember the basic trigonometric identities, especially the definitions of secant and tangent, and the Pythagorean identity!

STEP 2

1. Rewrite the equation
2. Multiply by a special form of one
3. Apply the Pythagorean identity
4. Simplify the expression

STEP 3

Let's **start** by writing down what we want to prove: cos⁡x1−sin⁡x=sec⁡x+tan⁡x \frac{\cos x}{1-\sin x} = \sec x + \tan x

STEP 4

We'll multiply the left side by a **clever form of one**, 1+sin⁡x1+sin⁡x\frac{1+\sin x}{1+\sin x}.
Why? Because it will help us simplify the denominator!
This is a valid operation because we're multiplying by one, which doesn't change the value of the expression. cos⁡x1−sin⁡x⋅1+sin⁡x1+sin⁡x \frac{\cos x}{1-\sin x} \cdot \frac{1+\sin x}{1+\sin x}

STEP 5

Now, let's **multiply** the numerators and denominators: cos⁡x(1+sin⁡x)(1−sin⁡x)(1+sin⁡x)=cos⁡x(1+sin⁡x)1−sin⁡2x \frac{\cos x (1+\sin x)}{(1-\sin x)(1+\sin x)} = \frac{\cos x (1+\sin x)}{1 - \sin^2 x} Notice how the denominator looks like a difference of squares!

STEP 6

Remember the **Pythagorean identity**: sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.
We can rewrite this as cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x.
This is **exactly** what we have in our denominator!
So let's **substitute**: cos⁡x(1+sin⁡x)1−sin⁡2x=cos⁡x(1+sin⁡x)cos⁡2x \frac{\cos x (1+\sin x)}{1 - \sin^2 x} = \frac{\cos x (1+\sin x)}{\cos^2 x}

STEP 7

We can **simplify** by dividing both the numerator and the denominator by cos⁡x\cos x.
Remember that cos⁡xcos⁡x=1\frac{\cos x}{\cos x} = 1, and cos⁡2xcos⁡x=cos⁡x\frac{\cos^2 x}{\cos x} = \cos x. cos⁡x(1+sin⁡x)cos⁡2x=1+sin⁡xcos⁡x \frac{\cos x (1+\sin x)}{\cos^2 x} = \frac{1+\sin x}{\cos x}

STEP 8

Now, we can **split** the fraction into two parts: 1+sin⁡xcos⁡x=1cos⁡x+sin⁡xcos⁡x \frac{1+\sin x}{\cos x} = \frac{1}{\cos x} + \frac{\sin x}{\cos x}

STEP 9

Finally, recall that sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} and tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}.
So, we have: 1cos⁡x+sin⁡xcos⁡x=sec⁡x+tan⁡x \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x

STEP 10

We've successfully shown that cos⁡x1−sin⁡x=sec⁡x+tan⁡x\frac{\cos x}{1-\sin x} = \sec x + \tan x!

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