Question

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

Answer 1

we know, $\cos 2A=2{\cos }^{2}A-1=1-2{\sin }^{2}A$

${\cos }^{2}A=\frac{1+\cos 2A}{2}$

${\sin }^{2}A=\frac{1-cos2A}{2}$

${\tan }^{2}A=\frac{si{n}^{2}A}{{\cos }^{2}A}=\frac{1-\cos 2A}{1+\cos 2A}$

Similarly, ${\tan }^{2}B=\frac{si{n}^{2}B}{{\cos }^{2}B}=\frac{1-\cos 2B}{1+\cos 2B}$

${\tan }^{2}A-{\tan }^{2}B=\frac{1-\cos 2A}{1+\cos 2A}-\frac{1-\cos 2B}{1+\cos 2B}$

add and substract 1

${\tan }^{2}A-{\tan }^{2}B=\frac{1-\cos 2A}{1+\cos 2A}+1-\frac{1-\cos 2B}{1+\cos 2B}-1$

$=(\frac{1-\cos 2A}{1+\cos 2A}+1)-(\frac{1-\cos 2B}{1+\cos 2B}+1)$

$=\left(\frac{2}{1+\cos 2A}\right)-\left(\frac{2}{1+\cos 2B}\right)=\left(\frac{2(\cos 2B-\cos 2A)}{(1+\cos 2B)(1+\cos 2A)}\right)$

$\cos 2A=1-2{\sin }^{2}A,\cos 2B=1-2{\sin }^{2}B$

${\tan }^{2}A-{\tan }^{2}B=\left(\frac{2(\cos 2B-\cos 2A)}{(1+\cos 2B)(1+\cos 2A)}\right)=\left(\frac{4({\sin }^{2}A-{\sin }^{2}B)}{(1+\cos 2B)(1+\cos 2A)}\right)$

$\cos 2A+1=2{\cos }^{2}A,\cos 2B+1=2{\cos }^{2}B$

${\tan }^{2}A-{\tan }^{2}B=\left(\frac{4({\sin }^{2}A-{\sin }^{2}B)}{(1+\cos 2B)(1+\cos 2A)}\right)=\left(\frac{4({\sin }^{2}A-{\sin }^{2}B)}{4{\cos }^{2}A{\cos }^{2}B}\right)=\left(\frac{{\sin }^{2}A-{\sin }^{2}B}{{\cos }^{2}A{\cos }^{2}B}\right)$