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 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

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})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

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

Suggest Corrections

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}

Join BYJU'S Learning Program

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

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