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869

Q:

$$ \sqrt {sec^2{θ}+cosec^2{θ}}=?$$

  • 1
    $$ {tan^2{θ}}+cot^2{θ}$$
  • 2
    $$ tanθ+cotθ$$
  • 3
    $$ sinθ+cosθ$$
  • 4
    None of these
  • Show Answer
  • Workspace

Answer : 2. "$$ tanθ+cotθ$$"
Explanation :

$$ \sqrt {sec^2{θ}+cosec^2{θ}}$$
We know that
$$ {1+tan^2{θ}}={sec^2θ}$$
$$ {1+cot^2{θ}}={cosec^2{θ}}$$
$$ \sqrt {1+tan^2{θ}+1+cot^2{θ}}$$

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