Question

If $a=tan\theta +cot\theta$ and $b=tan\theta –cot\theta$, then the value of $a^2-b^2+\left(\frac{a+b}{tan\theta }\right)$is equal to

• A.
  4
• B.
  $4tan\theta$
• C.
  1
• D.
  6

Answer 1

The correct option is D

6
Given: $a=tan\theta +cot\theta andb=tan\theta -cot\theta Consider,a^2-b^2+\left(\frac{a+b}{tan\theta }\right)=(tan\theta +cot\theta {)}^2-(tan\theta -cot\theta {)}^2+\left(\frac{tan\theta +cot\theta +tan\theta -cot\theta }{tan\theta }\right)=(ta{n}^2\theta +co{t}^2\theta +2tan\theta cot\theta )-(ta{n}^2\theta -co{t}^2\theta -2tan\theta cot\theta )+\left(\frac{2tan\theta }{tan\theta }\right)=ta{n}^2\theta +co{t}^2\theta +2tan\theta cot\theta -ta{n}^{\theta }-co{t}^2\theta +2tan\theta cot\theta +2=4tan\theta cot\theta +2=4tan\theta \times \frac{1}{tan\theta }+2=4+2=6$

Hence, the correct answer is option d.