Question

Prove the following identities:

${\displaystyle {\tan }^{2}A-{\tan }^{2}B=\frac{{\sin }^{2}A-{\sin }^{2}B}{{\cos }^{2}A\cdot {\cos }^{2}B}}$

Answer 1

L.H.S. = ${\displaystyle {\tan }^{2}A-{\tan }^{2}B}$

= ${\displaystyle \frac{{\sin }^{2}A}{{\cos }^{2}A}-\frac{{\sin }^{2}B}{{\cos }^{2}B}}$

= ${\displaystyle \frac{{\sin }^{2}A\cdot {\cos }^{2}B-{\sin }^{2}B\cdot {\cos }^{2}A}{{\cos }^{2}A\cdot {\cos }^{2}B}}$

= ${\displaystyle \frac{{\sin }^{2}A(1-{\sin }^{2}B)-{\sin }^{2}B(1-{\sin }^{2}A)}{{\cos }^{2}A\cdot {\cos }^{2}B}}$

= ${\displaystyle \frac{{\sin }^{2}A-{\sin }^{2}A\cdot {\sin }^{2}B-{\sin }^{2}B+{\sin }^{2}A\cdot {\sin }^{2}B}{{\cos }^{2}A\cdot {\cos }^{2}B}}$

= ${\displaystyle \frac{{\sin }^{2}A-\cancel{{\sin }^{2}A\cdot {\sin }^{2}B}-{\sin }^{2}B+\cancel{{\sin }^{2}A\cdot {\sin }^{2}B}}{{\cos }^{2}A\cdot {\cos }^{2}B}}$

= ${\displaystyle \frac{{\sin }^{2}A-{\sin }^{2}B}{{\cos }^{2}A\cdot {\cos }^{2}B}}$ = R.H.S.