How do you prove #\frac { \cos A + \sin A } { \cos A - \sin A } = an ( 45^ { \circ } + A )#?

Question

How do you prove #\frac { \cos A + \sin A } { \cos A - \sin A } = 	an ( 45^ { \circ } + A )#?

Violet Anderson · Accepted Answer

#LHS=(cosA+sinA)/(cosA-sinA)# #=((cosA+sinA)/cosA)/((cosA-sinA)/cosA)# #=(1+tanA)/(1-tanA)# #=(tan45°+tanA)/(1-tan45°*tanA)# #=tan(45°+A)=RHS#

Charles Autterson · Answer

undefined To prove $ \frac{\cos A + \sin A}{\cos A - \sin A} = \tan(45^\circ + A) $, we can start with the right-hand side and express it in terms of sine and cosine using trigonometric identities. Then, simplify and manipulate both sides until they are equal.

Using the angle addition formula for tangent, we have:

$$ \tan(45^\circ + A) = \frac{\sin(45^\circ + A)}{\cos(45^\circ + A)} $$

Applying the angle addition formulas for sine and cosine:

$$ \sin(45^\circ + A) = \sin 45^\circ \cos A + \cos 45^\circ \sin A = \frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A $$

$$ \cos(45^\circ + A) = \cos 45^\circ \cos A - \sin 45^\circ \sin A = \frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A $$

Substitute these expressions into the tangent expression:

$$ \tan(45^\circ + A) = \frac{\frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A}{\frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A} $$

$$ = \frac{\frac{1}{\sqrt{2}}(\cos A + \sin A)}{\frac{1}{\sqrt{2}}(\cos A - \sin A)} $$

$$ = \frac{\cos A + \sin A}{\cos A - \sin A} $$

Thus, we have shown that $ \frac{\cos A + \sin A}{\cos A - \sin A} = \tan(45^\circ + A) $.

How do you prove #\frac { \cos A + \sin A } { \cos A - \sin A } = \tan ( 45^ { \circ } + A )#?