Question

oliver incorrectly states that the expression \\( \tan \left( \frac { 3 \pi } { 4 } + x \
ight) \\) can be simplified as -1. review olivers work. \\( \tan \left( \frac { 3 \pi } { 4 } + x \
ight) \\) \\( = \frac { \tan \left( \frac { 3 \pi } { 4 } \
ight) + \tan ( x ) } { 1 - \tan \left( \frac { 3 \pi } { 4 } \
ight) \tan ( x ) } \\) \\( = \frac { - 1 + \tan ( x ) } { 1 - ( - 1 ) \tan ( x ) } \\) \\( = \frac { - 1 + \tan ( x ) } { 1 + \tan ( x ) } \\) \\( = - 1 \\) which statement explains why oliver is incorrect \\( \bigcirc \\) the expression \\( \frac { - 1 + \tan ( x ) } { 1 + \tan ( x ) } \\) does not simplify to -1. \\( \bigcirc \\) the expression \\( \tan \left( \frac { 3 \pi } { 4 } \
ight) \\) does not have a value of -1. \\( \bigcirc \\) the expression \\( 1 - ( - 1 ) \tan ( x ) \\) simplifies to \\( 2 \tan ( x ) \\), not \\( 1 + \tan ( x ) \\). \\( \bigcirc \\) the expression \\( \tan \left( \frac { 3 \pi } { 4 } + x \
ight) \\) is equivalent to \\( \tan \left( \frac { 3 \pi } { 4 } \
ight) + \tan ( x ) \\), not \\( \frac { \tan \left( \frac { 3 \pi } { 4 } \
ight) + \tan ( x ) } { 1 - \tan \left( \frac { 3 \pi } { 4 } \
ight) \tan ( x ) } \\)

Explanation:

Step1: Apply the tangent addition formula

The formula for \(\tan(A + B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}\). Here \(A=\frac{3\pi}{4}\) and \(B = x\), so \(\tan(\frac{3\pi}{4}+x)=\frac{\tan(\frac{3\pi}{4})+\tan(x)}{1-\tan(\frac{3\pi}{4})\tan(x)}\).

Step2: Calculate \(\tan(\frac{3\pi}{4})\)

We know that \(\tan(\frac{3\pi}{4})=- 1\). Substituting this value into the formula, we get \(\frac{-1+\tan(x)}{1-(-1)\tan(x)}=\frac{-1+\tan(x)}{1 + \tan(x)}\), which does not simplify to \(-1\).

Answer:

The expression \(\frac{-1+\tan(x)}{1+\tan(x)}\) does not simplify to \(-1\).