Q175 / 270 · NDA 2021 (Phase 1) · Mathematics

What is $\frac{1+\tan ^2 \theta}{1+\cot ^2 \theta}-\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^2$ equal to?

  • Correct answer
    A
    $0$
  • B
    $1$
  • C
    $2 \tan \theta$
  • D
    $2 \cot \theta$

Solution

$\frac{1+\tan ^2 \theta}{1+\cot ^2 \theta}-\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^2$ $\begin{aligned}& =\frac{1+\tan ^2 \theta}{\frac{\tan ^2 \theta+1}{\tan ^2 \theta}}-\left(\frac{1-\tan \theta}{\frac{\tan \theta-1}{\tan \theta}}\right)^2 \\& =\tan ^2 \theta-(-\tan \theta)^2=0 . \end{aligned}$

Hence, the correct answer is option: 'A'.

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