Question

Prove that :
$ta{n}^{2}A-ta{n}^{2}B=\frac{co{s}^{2}B-co{s}^{2}A}{co{s}^{2}Bco{s}^{2}A}=\frac{si{n}^{2}A-si{n}^{2}B}{co{s}^{2}Aco{s}^{2}B}$

Answer 1

We have,
L.H.S. = $ta{n}^{2}A-ta{n}^{2}B$

$\Rightarrow$ L.H.S. = $\frac{si{n}^{2}A}{co{s}^{2}A}-\frac{si{n}^{2}B}{co{s}^{2}B}$

$\Rightarrow$ L.H.S. = $\frac{si{n}^{2}Aco{s}^{2}B-co{s}^{2}Asi{n}^{2}B}{co{s}^{2}Aco{s}^{2}B}$

$\Rightarrow$ L.H.S. = $\frac{(1-co{s}^{2}A)co{s}^{2}B-co{s}^{2}A(1-co{s}^{2}B)}{co{s}^{2}Aco{s}^{2}B}$

$\Rightarrow$ L.H.S. = $\frac{co{s}^{2}B-co{s}^{2}Aco{s}^{2}B-co{s}^{2}A+co{s}^{2}Aco{s}^{2}B}{co{s}^{2}Aco{s}^{2}B}$

$\Rightarrow$ L.H.S. = $\frac{co{s}^{2}B-co{s}^{2}A}{co{s}^{2}Aco{s}^{2}B}$

$\Rightarrow$ L.H.S. = $\frac{(1-si{n}^{2}B)-(1-si{n}^{2}A)}{co{s}^{2}Aco{s}^{2}B}=\frac{si{n}^{2}A-si{n}^{2}B}{co{s}^{2}Aco{s}^{2}B}=R.H.S.$