2 tan -11-3-tan -11-7-

The correct option is D π/42tan^ 1 ( 1/3 )+tan^ 1 ( 1/7 )= =tan^ 12 ( 1/3 )/1 ( 1/3 )^2+tan^ 1 ( 1/7 ) =tan^ 1 ( 3/4 )+tan^ 1 ( 1/7 ) =tan^ 13/4+1/7/1 ( 3/4 ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

2 tan -11/3+t...

Question

$2 t a n^{- 1} (\frac{1}{3}) + t a n^{- 1} (\frac{1}{7}) =$

t a n^{- 1} (\frac{49}{29})

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\frac{π}{2}

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0

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\frac{π}{4}

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Solution

The correct option is D $\frac{π}{4}$
$2 t a n^{- 1} (\frac{1}{3}) + t a n^{- 1} (\frac{1}{7}) =$
$= t a n^{- 1} \frac{2 (\frac{1}{3})}{1 - {(\frac{1}{3})}^{2}} + t a n^{- 1} (\frac{1}{7})$
$= t a n^{- 1} (\frac{3}{4}) + t a n^{- 1} (\frac{1}{7})$
$= t a n^{- 1} \frac{\frac{3}{4} + \frac{1}{7}}{1 - (\frac{3}{4}) (\frac{1}{7})} = t a n^{- 1} 1 = \frac{π}{4}$

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Similar questions

Evaluate:

(a)

{sin}^{- 1} \frac{4}{5} + 2 {tan}^{- 1} \frac{1}{3} = \frac{π}{2}

(b)

{tan}^{- 1} \frac{1}{7} + 2 {tan}^{- 1} \frac{1}{3} = \frac{π}{4}

(c)

{tan}^{- 1} \frac{1}{5} + {tan}^{- 1} \frac{1}{7} + {tan}^{- 1} \frac{1}{3} + {tan}^{- 1} \frac{1}{8} = \frac{π}{4}

2 t a n^{- 1} (\frac{1}{3}) + t a n^{- 1} (\frac{1}{7}) = \frac{π}{4}

2 t a n^{- 1} (\frac{1}{3}) + t a n^{- 1} (\frac{1}{7})

is equal to

Q. The value of

2 {tan}^{- 1} \frac{1}{3} + {tan}^{- 1} \frac{1}{7}

Q. Prove:

2 {tan}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{1}{7} = {tan}^{- 1} \frac{31}{17}

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2tan−1(13)+tan−1(17)=

The correct option is D π42tan−1(13)+tan−1(17)= =tan−12(13)1−(13)2+tan−1(17) =tan−1(34)+tan−1(17) =tan−134+171−(34)(17)=tan−11=π4

$2 t a n^{- 1} (\frac{1}{3}) + t a n^{- 1} (\frac{1}{7}) =$