If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A or not.
Let △ABC in which ∠B = 90°
cot(A) = AB/BC = 4/3
Let AB = 4k an BC = 3k, where k is a positive real number.
According to the Pythagorean theorem,
AC2 = AB2 + BC2
AC2 = (4k)2+(3k)2
AC2 = 16k2+9k2
AC2 = 25k2
AC = 5k
Now, apply the values corresponding to the ratios
tan(A) = BC/AB = 3/4
sin (A) = BC/AC = 3/5
cos (A) = AB/AC = 4/5
Now compare the left hand side(LHS) with right hand side(RHS)
L.H.S = \(\frac{1-tan^2A}{1+tan^2A}\) = \(\frac{1-(\frac{3}{4})^2}{1+(\frac{3}{4})^2}\) = \(\frac{1-\frac{9}{16}}{1+\frac{9}{16}}\) = \(\frac{7}{25}\)
R.H.S = \(cos^2A - sin^2A\) = \((\frac{4}{5})^2- (\frac{3}{5})^2\) = \(\frac{16}{25}-\frac{9}{25}\) = \(\frac{7}{25}\)
Since, both the LHS and RHS = 7/25
R.H.S. =L.H.S.
Hence, (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A is proved