Join Examsbook
Answer : 2. "$$ tanθ+cotθ$$"
$$ \sqrt {sec^2{θ}+cosec^2{θ}}=?$$
5Q:
$$ \sqrt {sec^2{θ}+cosec^2{θ}}=?$$
- 1$$ {tan^2{θ}}+cot^2{θ}$$false
- 2$$ tanθ+cotθ$$true
- 3$$ sinθ+cosθ$$false
- 4None of thesefalse
- Show AnswerHide Answer
- Workspace
Answer : 2. "$$ tanθ+cotθ$$"
Explanation :
$$ \sqrt {sec^2{θ}+cosec^2{θ}}$$
We know that
$$ {1+tan^2{θ}}={sec^2θ}$$
$$ {1+cot^2{θ}}={cosec^2{θ}}$$
$$ \sqrt {1+tan^2{θ}+1+cot^2{θ}}$$