If `cos27^(@)=x` , the value of `tan63^(@)` is

Options

A. `(x)/(sqrt(1-x^(2)))`
B. `(x)/(sqrt(1+x^(2)))`
C. `(sqrt(1-x^(2)))/(x)`
D. `(sqrt(1+x^(2)))/(x)`

Answer

To solve the problem where \( \cos 27^\circ = x \) and we need to find the value of \( \tan 63^\circ \), we can follow these steps:

### Step 1: Understand the relationship between the angles
We know that \( 63^\circ \) is complementary to \( 27^\circ \) because:
\[ 63^\circ + 27^\circ = 90^\circ \]
This means that:
\[ \tan 63^\circ = \cot 27^\circ \]

### Step 2: Use the relationship between tangent and cosine
The cotangent of an angle is the reciprocal of the tangent:
\[ \cot \theta = \frac{1}{\tan \theta} \]
Thus, we can express \( \tan 63^\circ \) in terms of \( \cos 27^\circ \):
\[ \tan 63^\circ = \frac{1}{\tan 27^\circ} \]

### Step 3: Find \( \tan 27^\circ \) using cosine
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ \tan 27^\circ = \frac{\sin 27^\circ}{\cos 27^\circ} \]
Since we know \( \cos 27^\circ = x \), we need to find \( \sin 27^\circ \). Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
we can find \( \sin 27^\circ \):
\[ \sin^2 27^\circ = 1 - \cos^2 27^\circ = 1 - x^2 \]
Thus:
\[ \sin 27^\circ = \sqrt{1 - x^2} \]

### Step 4: Substitute back to find \( \tan 27^\circ \)
Now we can substitute \( \sin 27^\circ \) into the equation for \( \tan 27^\circ \):
\[ \tan 27^\circ = \frac{\sqrt{1 - x^2}}{x} \]

### Step 5: Find \( \tan 63^\circ \)
Now substituting \( \tan 27^\circ \) into the equation for \( \tan 63^\circ \):
\[ \tan 63^\circ = \frac{1}{\tan 27^\circ} = \frac{x}{\sqrt{1 - x^2}} \]

### Final Result
Thus, the value of \( \tan 63^\circ \) is:
\[ \tan 63^\circ = \frac{x}{\sqrt{1 - x^2}} \]