Question

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

Answer 1

LHS = tan²A - tan²B
= $\frac{si{n}^{2}A}{co{s}^{2}A}$ - $\frac{si{n}^{2}B}{co{s}^{2}B}$
= $\frac{si{n}^{2}A.co{s}^{2}B-si{n}^{2}B.co{s}^{2}A}{co{s}^{2}A.co{s}^{2}B}$
we know,
sin²x + cos²x = 1
So,cos²B = 1 - sin²B
cos²A = 1 - sin²A
Now substitute the value of cos²B and cos²A in the expression,

= $\frac{si{n}^{2}A(1-si{n}^{2}B)-si{n}^{2}B(1-si{n}^{2}A)}{co{s}^{2}A.co{s}^{2}B}$
= $\frac{si{n}^{2}A-si{n}^{2}A.si{n}^{2}B-si{n}^{2}B+si{n}^{2}A.si{n}^{2}B}{co{s}^{2}A.co{s}^{2}B}$
= $\frac{si{n}^{2}A-si{n}^{2}B}{co{s}^{2}A.co{s}^{2}B}$ = RHS