If 0-90- then - 2-cosec2-

The correct option is A tan θ + cot θ √(sec^2θ + cosec^2θ) =√((1 + tan^2θ) + (1 + cot^2θ)) =√(tan^2θ+cot^2θ + 2) =√(tan^2θ + cot^2θ + 2 tanθ cotθ) ...

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Byju's Answer

Standard IX

Mathematics

Trigonometric Identities

If 0∘<θ<90∘, ...

Question

If $0^{\circ} < θ < 90^{\circ}$ , then $\sqrt{s e c^{2} θ + c o s e c^{2} θ}$ =

tan θ + cot θ

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tan θ - cot θ

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- tan θ + cot θ

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- tan θ - cot θ

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Solution

The correct option is A
tan θ + cot θ

$\sqrt{s e c^{2} θ + c o s e c^{2} θ}$
$= \sqrt{(1 + t a n^{2} θ) + (1 + c o t^{2} θ)}$
$= \sqrt{t a n^{2} θ + c o t^{2} θ + 2}$
$= \sqrt{t a n^{2} θ + c o t^{2} θ + 2 t a n θ c o t θ}$ $[∵ t a n θ c o t θ = 1]$
$= \sqrt{(t a n θ + c o t θ)^{2}}$
$= t a n θ + c o t θ$

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If 0∘<θ<90∘, then √sec2θ+cosec2θ =

The correct option is A tan θ + cot θ √sec2θ+cosec2θ =√(1+tan2θ)+(1+cot2θ) =√tan2θ+cot2θ+2 =√tan2θ+cot2θ+2tanθcotθ [∵tanθcotθ=1] =√(tanθ+cotθ)2 =tanθ+cotθ

If $0^{\circ} < θ < 90^{\circ}$ , then $\sqrt{s e c^{2} θ + c o s e c^{2} θ}$ =