Sec2 - cot-1 1-2 -cosec2 - tan-1 1-3 -

^2 [ cot^ 1 ( 1/2 ) ]+cosec^2 ( tan^ 1 ( 1/3 ) ) =1+tan^2 ( cot^ 1 ( 1/2 ) )+cot^2 ( tan^ 1 ( 1/3 ) ) =1+[tan(tan^ 12)]^2+1+[cot(cot^ 13)]^2 =2+2^2+3^2 =15 ...

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Byju's Answer

Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

sec2 [ cot-1 ...

Question

$s e c^{2} [c o t^{- 1} (\frac{1}{2})] + c o s e c^{2} [t a n^{- 1} (\frac{1}{3})] =$

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Solution

The correct option is B
15

${sec}^{2} [c o t^{- 1} (\frac{1}{2})] + c o s e c^{2} (t a n^{- 1} (\frac{1}{3}))$
$= 1 + t a n^{2} (c o t^{- 1} (\frac{1}{2})) + c o t^{2} (t a n^{- 1} (\frac{1}{3}))$
$= 1 + [t a n (t a n^{- 1} 2)]^{2} + 1 + [c o t (c o t^{- 1} 3)]^{2}$
$= 2 + 2^{2} + 3^{2}$
$= 15$

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sec2[cot−1(12)]+cosec2[tan−1(13)]=

The correct option is B 15 sec2[cot−1(12)]+cosec2(tan−1(13)) =1+tan2(cot−1(12))+cot2(tan−1(13)) =1+[tan(tan−12)]2+1+[cot(cot−13)]2 =2+22+32 =15

$s e c^{2} [c o t^{- 1} (\frac{1}{2})] + c o s e c^{2} [t a n^{- 1} (\frac{1}{3})] =$

The correct option is B
15

${sec}^{2} [c o t^{- 1} (\frac{1}{2})] + c o s e c^{2} (t a n^{- 1} (\frac{1}{3}))$
$= 1 + t a n^{2} (c o t^{- 1} (\frac{1}{2})) + c o t^{2} (t a n^{- 1} (\frac{1}{3}))$
$= 1 + [t a n (t a n^{- 1} 2)]^{2} + 1 + [c o t (c o t^{- 1} 3)]^{2}$
$= 2 + 2^{2} + 3^{2}$
$= 15$