RHS $\displaystyle \frac{{\tan x-1}}{{\tan x+1}}$

Change tan x  to sin x /cos x  and 1 to cos x /cos x. $\displaystyle \frac{{\frac{{\sin x}}{{\cos x}}-\frac{{\cos x}}{{\cos x}}}}{{\frac{{\sin x}}{{\cos x}}+\frac{{\cos x}}{{\cos x}}}}$

Combine the numerator to a single fraction, do the same for the denominator. $\displaystyle \frac{{\frac{{\sin x-\cos x}}{{\cos x}}}}{{\frac{{\sin x+\cos x}}{{\cos x}}}}$

Eliminate cos x in the denominators. $\displaystyle \frac{{\sin x-\cos x}}{{\sin x+\cos x}}$

Multiply both the numerator and denominator by sin x – cos x. $\displaystyle \frac{{\sin x-\cos x}}{{\sin x+\cos x}}\times \frac{{\sin x-\cos x}}{{\sin x-\cos x}}$

Expand the numerator and simplify $\displaystyle \frac{{(\sin x-\cos x)(\sin x-\cos x)}}{{{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}$ $\displaystyle \frac{{{{{\sin }}^{2}}x-2\sin x\cos x+{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}$ $\displaystyle \frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x-2\sin x\cos x}}{{{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}$

Convert sin2x + cos2x =1 . $\displaystyle \frac{{1-2\sin x\cos x}}{{{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}$

LHS = RHS     Shown

You are told in school to start proving from the more complicated side, but if you look at the question above, both sides seems equally complicated. So how do we determine which side to start with? I choose the right-hand side (RHS)  as it is easier to convert tan to sin and cos rather than the other way round.  Always keep the exam formula list beside you when you do proving, not only will you be able to do the question faster as you don’t have to recall from memory, the  trigonometric formulas may also provide clues on how to carry out the proving.

Small Group Tuition Additional Math (A-Math) at Woodlands, Elementary Math (E-Math) at Admiralty, Woodlands and Johor Bahru.