Question

Prove that (1 + tan 1°)(1 + tan 2°)(1 + tan 3°) ..... (1 + tan 44°) is a multiple of 4

Answer 1

1 + tan 44° = 1 + tan(45° – 1°)

= ${\displaystyle 1+\frac{{\tan 45}^{\circ }-{\tan 1}^{\circ }}{1-{\tan 45}^{\circ }{\tan 1}^{\circ }}}$

= ${\displaystyle 1+\frac{1-{\tan 1}^{\circ }}{1+{\tan 1}^{\circ }}}$

= ${\displaystyle \frac{1+{\tan 1}^{\circ }+1-{\tan 1}^{\circ }}{1-1{\tan 1}^{\circ }}}$

= ${\displaystyle \frac{2}{1-1{\tan 1}^{\circ }}}$

(1 + tan 1°)(1 + tan 44°) = 2

Similarly (1 + tan 2°)(1 + tan 43°) = 2

(1 + tan 3°)(1 + tan 42°) = 2

(1 + tan 22°)(1 + tan 23°) = 2

= (1 + tan 1°)(1 + tan 2°) … (1 + tan 44°)

= 2 × 2 × … 22 times

It is a multiple of 4.