If `cos x=0.32`, then what is the value of `sinx tanx`?

Answer

To find the value of \( \sin x \tan x \) given that \( \cos x = 0.32 \), we can follow these steps:

### Step 1: Calculate \( \sin x \)
We know the relationship between sine and cosine:
\[ \sin^2 x + \cos^2 x = 1 \]
From this, we can express \( \sin x \) as:
\[ \sin x = \sqrt{1 - \cos^2 x} \]
Substituting the value of \( \cos x \):
\[ \sin x = \sqrt{1 - (0.32)^2} \]
Calculating \( (0.32)^2 \):
\[ (0.32)^2 = 0.1024 \]
Now substituting this back:
\[ \sin x = \sqrt{1 - 0.1024} = \sqrt{0.8976} \]
Calculating \( \sqrt{0.8976} \):
\[ \sin x \approx 0.9474 \]

### Step 2: Calculate \( \tan x \)
The tangent function is defined as:
\[ \tan x = \frac{\sin x}{\cos x} \]
Substituting the values we have:
\[ \tan x = \frac{0.9474}{0.32} \]
Calculating this gives:
\[ \tan x \approx 2.9606 \]

### Step 3: Calculate \( \sin x \tan x \)
Now we can find \( \sin x \tan x \):
\[ \sin x \tan x = 0.9474 \times 2.9606 \]
Calculating this product:
\[ \sin x \tan x \approx 2.81 \]

### Final Answer
Thus, the value of \( \sin x \tan x \) is approximately \( 2.81 \).

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