Question

$\frac{(1+tan13)(1+tan32)}{1+tan12\left)\right(1+tan33)}=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

$\frac{(1+\tan 13)(1+\tan 32)}{(1+\tan 12)(1+\tan 33)}$

$\frac{(1+\tan 13)(1+\tan (45-13\left)\right)}{(1+\tan 12)(1+\tan (45-12\left)\right)}$

$\frac{(1+\tan 13)(1+\frac{1-\tan 13}{1+\tan 13})}{(1+\tan 12)(1+\frac{1-\tan 12}{1+\tan 12})}$

$\left\{\begin{array}{c}\tan (45-\theta )\\ =\frac{1-\tan \theta }{1+\tan \theta }\end{array}\right\}$

$\frac{(1+\frac{\tan 13)}{(1+\tan 13)}+1-\tan 13)}{\frac{1+\tan 13+1-\tan }{(1+\tan 13)}}$

$\frac{(1+\tan 12)(1+\tan 12+1-\tan 12)}{(1+\tan 12)}$

$\begin{array}{cc}=& \frac{1+1}{1+1}=\frac{2}{2}=1\\ & \text{option}A\text{is correct.}\end{array}$