given: \\( \tan a = - sqrt { 15 } \\)\nwhat is the value of \\( \tan left( a - \frac { pi } { 4 } \right)…

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

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

Answer

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)