Question

given: \\( \tan a = - sqrt { 15 } \\)
what is the value of \\( \tan left( a - \frac { pi } { 4 }
ight) \\)?
\\( \frac { sqrt { 15 } + 1 } { 1 - sqrt { 15 } } \\)
\\( \frac { - sqrt { 15 } + 1 } { 1 + sqrt { 15 } } \\)
\\( \frac { sqrt { 15 } - 1 } { 1 + sqrt { 15 } } \\)
\\( \frac { - sqrt { 15 } - 1 } { 1 - sqrt { 15 } } \\)

Explanation:

Step1: Recall the tangent subtraction formula

The formula for \(\tan(x - y)=\frac{\tan x-\tan y}{1 + \tan x\tan y}\). Here \(x = A\) and \(y=\frac{\pi}{4}\), and \(\tan\frac{\pi}{4}=1\).

Step2: Substitute the values into the formula

Substitute \(\tan A=-\sqrt{15}\) and \(\tan\frac{\pi}{4} = 1\) into \(\tan(A-\frac{\pi}{4})=\frac{\tan A-\tan\frac{\pi}{4}}{1+\tan A\tan\frac{\pi}{4}}\).
We get \(\tan(A - \frac{\pi}{4})=\frac{-\sqrt{15}-1}{1+(-\sqrt{15})\times1}=\frac{-\sqrt{15}-1}{1-\sqrt{15}}\).

Answer:

\(\frac{-\sqrt{15}-1}{1 - \sqrt{15}}\) (corresponding to the fourth option)