2tan -1 2-cosec2 -1 3-

The correct option is B 15Let tan^ 1 2 = α⇒ tan α = 2 and cot^ 1 3 = β⇒ cot β = 3 sec^2 (tan^ 1 2) + cosec^2 (cot^ 13) = sec^2 α + cosec^2 β = 1 + tan^2 α +1 ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

2tan -1 2+cos...

Question

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

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Solution

The correct option is C 15
Let $t a n^{- 1} 2 = α \Rightarrow t a n α = 2$
and $c o t^{- 1} 3 = β \Rightarrow c o t β = 3 s e c^{2} (t a n^{- 1} 2) + c o s e c^{2} (c o t^{- 1} 3) = s e c^{2} α + c o s e c^{2} β = 1 + t a n^{2} α + 1 + c o t^{2} β = 2 + (2)^{2} + (3)^{2} = 15.$

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Similar questions

Q. Evaluate:

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

Q. Prove that

{sec}^{2} ({tan}^{- 1} 2) + {cosec}^{2} ({cot}^{- 1} 3) = 15

Q. The value of

{sec}^{2} ({tan}^{- 1} 2) + {cosec}^{2} ({cot}^{- 1} 3)

Q. The value of

{sec}^{2} ({tan}^{- 1} 3) + {cosec}^{2} ({cot}^{- 1} 2)

is equal to

Q. If

x = {sec}^{2} ({tan}^{- 1} 2) + {cosec}^{2} ({cot}^{- 1} 3)

and

y = {tan}^{- 1} \frac{\sqrt{1 + {tan}^{2} 8} - 1}{tan 8}

then

15 x^{2} + 4 y^{2}

is equal to

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sec2(tan−12)+cosec2(cot−13)=

The correct option is C 15Let tan−1 2=α⇒tan α=2 and cot−1 3=β⇒cot β=3sec2(tan−12)+cosec2(cot−13)=sec2 α+cosec2 β=1+tan2 α+1+cot2 β=2+(2)2+(3)2=15.

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

The correct option is C 15
Let $t a n^{- 1} 2 = α \Rightarrow t a n α = 2$
and $c o t^{- 1} 3 = β \Rightarrow c o t β = 3 s e c^{2} (t a n^{- 1} 2) + c o s e c^{2} (c o t^{- 1} 3) = s e c^{2} α + c o s e c^{2} β = 1 + t a n^{2} α + 1 + c o t^{2} β = 2 + (2)^{2} + (3)^{2} = 15.$