Question

Given that cos x=12/13. Evaluate t-tan x/tan x

Answer 1

To find \( t - \frac{\tan x}{\tan x} \), let's first express \( t \).

Given \( \cos x = \frac{12}{13} \), we can find \( \sin x \) using the Pythagorean identity:

\[
\sin^2 x + \cos^2 x = 1
\]

Substituting \(\cos x\):

\[
\sin^2 x + \left(\frac{12}{13}\right)^2 = 1
\]
\[
\sin^2 x + \frac{144}{169} = 1
\]

Subtract \(\frac{144}{169}\) from both sides:

\[
\sin^2 x = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}
\]

Taking the square root:

\[
\sin x = \frac{5}{13} \quad (\text{since sine is positive in the first quadrant})
\]

Next, we calculate \(\tan x\):

\[
\tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}
\]

Now, we can evaluate \( t - \frac{\tan x}{\tan x} \).

Since \(\frac{\tan x}{\tan x} = 1\), we have:

\[
t - \frac{\tan x}{\tan x} = t - 1
\]

We do not have a specific value of \( t \) defined in the problem. Assuming \( t \) represents the tangent, then substituting \( t \) with \( \tan x \):

\[
\tan x = \frac{5}{12}
\]

Thus,

\[
\tan x - 1 = \frac{5}{12} - 1 = \frac{5}{12} - \frac{12}{12} = -\frac{7}{12}
\]

However, since \( t \) has not been explicitly defined, \( t - 1 \) cannot be evaluated numerically unless we assign a value to \( t \).

If \( t \) is simply left as \( t \), we present the final output as:

\[
t - 1
\]

If you have a specific value for \( t \), substitute that value accordingly. Otherwise, the expression simplifies to:

\[
t - 1
\]