Question

If cotA = $\frac{4}{3}$, then $\frac{(1-ta{n}^{2}A)}{(1+ta{n}^{2}A)}=co{s}^{2}A-si{n}^{2}A$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $\Delta$ABC in which $\angle B={90}^{\circ }$,
According to question,
$cotA=\frac{AB}{BC}=\frac{4}{3}$
Let AB = 4k and BC = 3k,where k is a positive real number.
By Pythagoras theorem in $\Delta$ABC we get.
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(4k{)}^2+(3k{)}^2$
$A{C}^2=16k^2+9k^2$
$A{C}^2=25k^2$
AC = 5k
$tanA=\frac{BC}{AB}=\frac{3}{4}$
$sinA=\frac{BC}{AC}=\frac{3}{5}$
$cosA=\frac{AB}{AC}=\frac{4}{5}$
$L.H.S=\frac{(1-ta{n}^{2}A)}{(1+ta{n}^{2}A)}$
$=\frac{1-{\left(\frac{3}{4}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}$
$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$
$=\frac{16-9}{16+9}=\frac{7}{25}$
$R.H.S=co{s}^{2}A-si{n}^{2}A$
$=(\frac{4}{5}{)}^2-(\frac{3}{4}{)}^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$
R.H.S = L.H.S
$Hence,\frac{(1-ta{n}^{2}A)}{(1+ta{n}^{2}A)}=co{s}^{2}A-si{n}^{2}A$