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3 years ago

Proprtaions with tables

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

8 0

Answer:

I graphed it, the first one is correct .

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option 1 drop down are: even-odd identity, quotient identity, Pythagorean identity, double-number identity.option 2 drop down ar

motikmotik

Answer:

The equation is given below as

$\frac{\cos2x}{\cos x}=\cos x-\sin x\tan x$

Step 1:

We will work on the left-hand side, we will have

$\begin{gathered} \cos x-\sin x\tan x \\ \text{recall that,} \\ Quoitent\text{ identity is} \\ \tan x=\frac{\sin x}{\cos x} \end{gathered}$

By substituting the identity above, we will have

$\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x}=\cos x-\frac{\sin^2x}{\cos x} \\ \end{gathered}$

Here, we will make use of the quotient identity

Step 2:

By writings an expression, we will have

$\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x} \\ \cos x-\sin x\tan x=\frac{\cos^2x-\sin^2x}{\cos x} \end{gathered}$

Here, we will use the definition of subtraction

$\cos x-\frac{\sin^2x}{\cos x}$

Step 3:

We will apply the double number identity given below

$\begin{gathered} \cos 2\theta=\cos (\theta+\theta)=\cos ^2\theta-\sin ^2\theta \\ \cos 2x=cos(x+x)=\cos ^2x-\sin ^2x \end{gathered}$

By applying this, we will have

$\frac{\cos^2x-\sin^2x}{\cos x}=\frac{\cos2x}{\cos x}$

Here, we will use the double number identity

$\frac{\cos^2x-\sin^2x}{\cos x}$

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