Question

Number of real solutions of the equation $(tanx+1)(tanx+3)(tanx+5)(tanx+7)=33$

• A. Will be 2 in the interval $[-\pi /2,\pi /2]$
• B. Will be 4 in the interval $[-\pi /2,\pi /2]$
• C. Will be 3 in the interval $[-\pi /2,\pi )$
• D. Will be 4 in the interval $[-\pi /2,\pi )$

Answer 1

Answer: (A), (D)

The correct options are
A Will be 2 in the interval $[-\pi /2,\pi /2]$
D Will be 4 in the interval $[-\pi /2,\pi )$

$(ta{n}^{2}x+8tanx+15)(ta{n}^{2}x+8tanx+7)=33$
Let $ta{n}^{2}x+8tanx=p$
$(p+15)(p+7)=33$
$\Rightarrow p^2+22p+72=0$
$\Rightarrow p=-18$ (rejected) or $p=-4$
$\Rightarrow ta{n}^{2}x+8tanx+4=0$
so, $tanx+4=\pm \sqrt{12}\Rightarrow tanx=-4+\sqrt{12}$ or $-4-\sqrt{12}$