Question

Prove the following identity....
$\frac{(ta{n}^{2}A-ta{n}^{2}B=(si{n}^{2}A-si{n}^{2}B)}{(Co{s}^{2}A\times Co{s}^{2}B)}$

Answer 1

$\frac{(ta{n}^{2}A-ta{n}^{2}B=(si{n}^{2}A-si{n}^{2}B)}{(Co{s}^{2}A.Co{s}^{2}B)}$
L.H.S =$ta{n}^{2}A-ta{n}^{2}B$
$\Rightarrow Useta{n}^{2}x=se{c}^{2}x-1$
$=se{c}^{2}A-1-se{c}^{2}B-1$
$=se{c}^{2}A-se{c}^{2}B$
$=\frac{1}{co{s}^{2}A}-\frac{1}{co{s}^{2}B}$
$=\frac{(co{s}^{2}B-co{s}^{2}A)}{co{s}^{2}A.co{s}^{2}B}$
$\Rightarrow$ use $co{s}^{2}x=1-si{n}^{2}x$
$=\frac{(1-si{n}^{2}B-1+si{n}^{2}A)}{co{s}^{2}A.co{s}^{2}B}$
$=\frac{(si{n}^{2}A-si{n}^{2}B)}{co{s}^{2}A.co{s}^{2}B}$
Hence L:H:S=R:H:S