Question

If $tan(\pi /4+\theta )+tan(\pi /4-\theta )=a^2$ then $ta{n}^2(\pi /4+\theta )+ta{n}^2(\pi /4-\theta )$

• A. $a^2+1$
  
• B. $a^2+2$
  
• C. $a^2-2$
  
• D. none of these

Answer 1

The correct option is C $a^2-2$


Given:
$tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta )=a$
$\Rightarrow {[tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta )]}^2=a^2$
$\Rightarrow ta{n}^2(\frac{\pi }{4}+\theta )+ta{n}^2(\frac{\pi }{4}-\theta )+2tan(\frac{\pi }{4}-\theta )tan(\frac{\pi }{4}+\theta )$
$\Rightarrow ta{n}^2(\frac{\pi }{4}+\theta )+ta{n}^2(\frac{\pi }{4}-\theta )$
$a^2-2tan[\frac{tan{45}^{\circ }-tan\theta }{1+tan{45}^{\circ }tan\theta }\times \frac{tan{45}^{\circ }+tan\theta }{1-tan{45}^{\circ }tan\theta }]$
$\Rightarrow ta{n}^2(\frac{\pi }{4}+\theta )+ta{n}^2(\frac{\pi }{4}-\theta )$
$=a^2-2[\frac{1-tan\theta }{1+tan\theta }\times \frac{1+tan\theta }{1-tan\theta }]$
$\Rightarrow ta{n}^2(\frac{\pi }{4}+\theta )+ta{n}^2(\frac{\pi }{4}-\theta )$
$=a^2-2\left(\frac{1-ta{n}^2\theta }{1-ta{n}^2\theta }\right)$
$\Rightarrow ta{n}^2(\frac{\pi }{4}+\theta )+ta{n}^2(\frac{\pi }{4}-\theta )=a^2-2$