The correct option is B sec^2θ+cosec^2θ tan^2θ+cot^2θ+2 This can be re written as nbsp; ⇒ 1 + nbsp; nbsp;tan^2θ+ 1 + cot^2θ ⇒ cosec^2 θ + sec ^2 θ (∵ n ...

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Trigonometric Identities

tan2θ+cot2θ+2...

Question

$t a n^{2} θ + c o t^{2} θ + 2 =$

$t a n^{2} θ + c o s e c^{2} θ$

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$s e c^{2} θ + c o s e c^{2} θ$

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s e c^{2} θ - c o s e c^{2} θ + 2

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s e c^{2} θ + c o s e c^{2} θ - 2

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Solution

The correct option is B
$s e c^{2} θ + c o s e c^{2} θ$

$t a n^{2} θ + c o t^{2} θ + 2$

This can be re-written as

$\Rightarrow 1 + t a n^{2} θ + 1 + c o t^{2} θ$

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

Hence
$t a n^{2} θ + c o t^{2} θ + 2 = s e c^{2} θ + c o s e c^{2} θ$

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Similar questions

t a n^{2} θ + c o t^{2} θ + 2 = s e c^{2} θ + c o s e c^{2} θ

Q. Prove that :

\frac{t a n θ - c o t θ}{s i n θ c o s θ} = s e c^{2} θ - c o s e c^{2} θ = t a n^{2} θ - c o t^{2} θ

Q. If θ is an acute angel such that

s e c^{2} θ = 3,

then find the value of

\frac{{tan}^{2} θ - c o s e c^{2} θ}{{tan}^{2} θ + c o s e c^{2} θ}

Q. Prove that

{sec}^{2} θ - cos e c^{2} θ = {tan}^{2} θ - {cot}^{2} θ

Q. Write the value of

\frac{\tan^{2} θ - \sec^{2} θ}{\cot^{2} θ - {cosec}^{2} θ}

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tan2θ+cot2θ+2=

The correct option is B sec2θ+cosec2θ tan2θ+cot2θ+2 This can be re-written as ⇒1+tan2θ+1+cot2θ ⇒cosec2θ+sec2θ (∵1+tan2θ=cosec2θ and 1+cot2θ=sec2θ) Hence tan2θ+cot2θ+2=sec2θ+cosec2θ

$t a n^{2} θ + c o t^{2} θ + 2 =$