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Math  /  Trigonometry

Question5. Prove the following identities a) cos⁡(β)+tan⁡(β)sin⁡(β)=sec⁡(β)\cos (\beta)+\tan (\beta) \sin (\beta)=\sec (\beta)

Studdy Solution

STEP 1

What is this asking? We need to show that adding cosine of an angle to the product of its tangent and sine results in the secant of that same angle. Watch out! Remember those trigonometric identities!
We'll need them here.
Also, be careful not to mix up sec⁡(β) \sec(\beta) and csc⁡(β) \csc(\beta) .

STEP 2

1. Rewrite everything in terms of sine and cosine.
2. Simplify the expression.

STEP 3

Alright, let's **rewrite** the left side of the equation in terms of sine and cosine!
Remember, tan⁡(β)=sin⁡(β)cos⁡(β) \tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} and sec⁡(β)=1cos⁡(β) \sec(\beta) = \frac{1}{\cos(\beta)} .
This is a super useful trick for simplifying trigonometric expressions.

STEP 4

So, we have: cos⁡(β)+tan⁡(β)sin⁡(β)=cos⁡(β)+sin⁡(β)cos⁡(β)⋅sin⁡(β) \cos(\beta) + \tan(\beta)\sin(\beta) = \cos(\beta) + \frac{\sin(\beta)}{\cos(\beta)} \cdot \sin(\beta) See how we swapped out tan⁡(β) \tan(\beta) for its equivalent in terms of sine and cosine?

STEP 5

Now, let's **simplify** that expression we just got.
Multiply those sines together in the second term: cos⁡(β)+sin⁡(β)cos⁡(β)⋅sin⁡(β)=cos⁡(β)+sin⁡2(β)cos⁡(β) \cos(\beta) + \frac{\sin(\beta)}{\cos(\beta)} \cdot \sin(\beta) = \cos(\beta) + \frac{\sin^2(\beta)}{\cos(\beta)}

STEP 6

To add these two terms together, we need a common denominator.
The common denominator is cos⁡(β) \cos(\beta) , so let's multiply the first term, cos⁡(β) \cos(\beta) , by cos⁡(β)cos⁡(β) \frac{\cos(\beta)}{\cos(\beta)} which is just like multiplying by **one**!
This gives us: cos⁡(β)⋅cos⁡(β)cos⁡(β)+sin⁡2(β)cos⁡(β)=cos⁡2(β)cos⁡(β)+sin⁡2(β)cos⁡(β) \cos(\beta) \cdot \frac{\cos(\beta)}{\cos(\beta)} + \frac{\sin^2(\beta)}{\cos(\beta)} = \frac{\cos^2(\beta)}{\cos(\beta)} + \frac{\sin^2(\beta)}{\cos(\beta)}

STEP 7

Now, with that common denominator, we can **add** the numerators: cos⁡2(β)cos⁡(β)+sin⁡2(β)cos⁡(β)=cos⁡2(β)+sin⁡2(β)cos⁡(β) \frac{\cos^2(\beta)}{\cos(\beta)} + \frac{\sin^2(\beta)}{\cos(\beta)} = \frac{\cos^2(\beta) + \sin^2(\beta)}{\cos(\beta)}

STEP 8

And here comes the *magic*!
Remember the **Pythagorean identity**: cos⁡2(β)+sin⁡2(β)=1 \cos^2(\beta) + \sin^2(\beta) = 1 !
So, we can **substitute** 1 for the numerator: cos⁡2(β)+sin⁡2(β)cos⁡(β)=1cos⁡(β) \frac{\cos^2(\beta) + \sin^2(\beta)}{\cos(\beta)} = \frac{1}{\cos(\beta)}

STEP 9

Finally, remember that 1cos⁡(β) \frac{1}{\cos(\beta)} is the same as sec⁡(β) \sec(\beta) ! 1cos⁡(β)=sec⁡(β) \frac{1}{\cos(\beta)} = \sec(\beta)

STEP 10

We've shown that cos⁡(β)+tan⁡(β)sin⁡(β)=sec⁡(β) \cos(\beta) + \tan(\beta)\sin(\beta) = \sec(\beta) , which is exactly what we wanted to prove!

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