Question

$\frac{(1+tanAtanB{)}^2+(tanA-tanB{)}^2}{{sec}^{2}A{sec}^{2}B}=$ ___

• A. $1$
• B. $\text{-tan A}$
• C. $2$
• D. $\text{cot A}$

Answer 1

The correct option is A

$1$

$\frac{(1+\text{tan A tan B}{)}^2+(\text{tan A-tan B}{)}^2}{{\text{sec}}^{2}A{\text{sec}}^{2}B}$

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

$=\frac{1+{\text{tan}}^{2}A{\text{tan}}^{2}B+{\text{tan}}^{2}A+{\text{tan}}^{2}B}{{\text{sec}}^{2}A{\text{sec}}^{2}B}$

$=\frac{1+{\text{tan}}^{2}A+{\text{tan}}^{2}B+{\text{tan}}^{2}A{\text{tan}}^{2}B}{{\text{sec}}^{2}A{\text{sec}}^{2}B}$

$=\frac{1(1+{\text{tan}}^{2}A)+{\text{tan}}^{2}B(1+{\text{tan}}^{2}A)}{{\text{sec}}^{2}A{\text{sec}}^{2}B}$

$=\frac{(1+{\text{tan}}^{2}A)(1+{\text{tan}}^{2}B)}{{\text{sec}}^{2}A{\text{sec}}^{2}B}$

$=\frac{\left({\text{sec}}^{2}A\right)\left({\text{sec}}^{2}B\right)}{{\text{sec}}^{2}A{\text{sec}}^{2}B}.\dots (1+{\text{tan}}^2\theta ={\text{sec}}^2\theta )$

$=1$