Question

Question 8
If cot A = $\frac{4}{3}$, check whether $\frac{(1-ta{n}^{2}A)}{(1+ta{n}^{2}A)}=co{s}^{2}A-si{n}^{2}A$ or not.

Answer 1

Let $\Delta$ABC in which $\angle B={90}^{\circ }$,
According to the 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. = $\frac{(1-ta{n}^{2}A)}{(1+ta{n}^{2}A)}$
= $\frac{1-{\frac{3}{4}}^2}{1+{\frac{3}{4}}^2}$=$\frac{(1-\frac{9}{16})}{(1+\frac{9}{16})}$= $\frac{(16-9)}{(16+9)}$ = $\frac{7}{25}$
R.H.S. = $co{s}^{2}A-si{n}^{2}A$ = $(\frac{4}{5}{)}^2-(\frac{3}{5}{)}^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$
R.H.S. = L.H.S.
Hence, $\frac{(1-ta{n}^{2}A)}{(1+ta{n}^{2}A)}=co{s}^{2}A-si{n}^{2}A$