MCQ
$\cot\left(45^{\circ}+\theta\right)\cot\left(45^{\circ}-\theta\right)=$
  • A
    -1
  • B
    $0$
  • 1
  • D
    $\infty$

Answer

Correct option: C.
1
(C)
$\cot\left(45^{\circ}+\theta\right)\cot\left(45^{\circ}-\theta\right)$
$=\tan\left(90^{\circ}-45^{\circ}-\theta\right)\cot\left(45^{\circ} \theta\right)$
$\ldots\left[\because\tan\left(90^{\circ}-\theta\right)=\cot\theta\right]$
$=\tan\left(45^{\circ}-\theta\right)\cot\left(45^{\circ}-\theta\right)=1$

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