$s e c^{2} (t a n^{- 1} 2) + c o s e c^{2} (c o t^{- 1} 3)$ is equal to

5

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13

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15

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6

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Solution

The correct option is C $15$
Given, ${sec}^{2} ({tan}^{- 1} (2)) + c o s e c^{2} ({cot}^{- 1} (3))$
$= 2 + {tan}^{2} ({tan}^{- 1} (2)) + {cot}^{2} ({cot}^{- 1} (3))$
$= 2 + 2^{2} + 3^{2}$
$= 2 + 4 + 9$
$= 15$

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Similar questions

s e c^{2} (t a n^{- 1} 2) + c o s e c^{2} (c o t^{- 1} 3)

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

Q. If

A = 13 [12 + {32 \times 5 + (32 - 12 \div 3)}]

and

B = 13 {- 15 + (13) \times 4}

, then

A \div B

is equal to

Q. Find the value of ;
(i)

{sec}^{2} ({tan}^{- 1} 2) + c o s e c^{2} (c o t^{- 1} 3) + 2

(ii)

tan ({sin}^{- 1} \frac{3}{5} + {cot}^{- 1} \frac{3}{2})

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

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sec2(tan−12)+cosec2(cot−13) is equal to

The correct option is C 15Given, sec2(tan−1(2))+cosec2(cot−1(3))=2+tan2(tan−1(2))+cot2(cot−1(3))=2+22+32=2+4+9=15

$s e c^{2} (t a n^{- 1} 2) + c o s e c^{2} (c o t^{- 1} 3)$ is equal to

The correct option is C $15$
Given, ${sec}^{2} ({tan}^{- 1} (2)) + c o s e c^{2} ({cot}^{- 1} (3))$
$= 2 + {tan}^{2} ({tan}^{- 1} (2)) + {cot}^{2} ({cot}^{- 1} (3))$
$= 2 + 2^{2} + 3^{2}$
$= 2 + 4 + 9$
$= 15$