Question

Find x,if 2tan^-1 (1/3)+tan^-1 (1/7) = tan^1 x
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ANSWER X=1

Answer 1

\[2\tan ^{-1} \frac{ 1 }{3 }+\tan ^{-1}\frac{ 1 }{ 7 }=\tan ^{-1}x\]

Answer 2

I am havin problem in end

Answer 3

Let\[\theta=\tan ^{-1}x\]
\[x=\tan \theta\]

Answer 4

x=tan[2tan^-1 (1/3) +tan^-1 (1/7)]

Answer 5

Let A = tan^-1 (1/3)
tanA=1/3

Answer 6

Let B = tan^-1 (1/7)
tanB=1/7

Answer 7

x=tan{tan2A+tanB}

Answer 8

\[As \tan(a-b) = \frac{ tana+tanb }{ 1-tana*tanb }\]

Answer 9

x = tan(2A + B) not x=tan{tan2A+tanB}

Answer 10

Glad you figured out your problem
:)

Answer 11

wait I am doing it

Answer 12

ok

Answer 13

what is the formula for tan2A ?

Answer 14

\[x=\frac{ \tan2A+tanB }{1-\tan2A*tanB }\]

Answer 15

we know these: tanA=1/3 and tanB=1/7
what is formula for calculating 2A

Answer 16

\[\tan (A + B) = \frac{ \tan A + \tan B }{ 1 - (\tan A \tan B) }\]

put A = B for tan 2A

Answer 17

yes I got it "the formula"

Answer 18

It says:

Answer 19

\[\tan2 \theta = \frac{ 2\tan \theta }{ 1-\tan^2 \theta }\]

Answer 20

yes thats right!!

Answer 21

tan2A = 2tanA/1-tan^2A

Answer 22

yes now you calculate tan2A and tell me ? or if you can then go for answer

Answer 23

ok

Answer 24

3/4

Answer 25

why don;t you simply mug up the formula :

tan^-1 a + tan^-1 b = tan^-1 (a+b)/(1-ab)

Answer 26

@shubhamsrg let @koli123able concentrate of concepts initially ....my suggestion other what @koli123able feels fine we will tell accordingly

Answer 27

I got it x=1

Answer 28

cool so main mistake was writing x=tan{tan2A+tanB} instead of x = tan(2A + B)
so remember this part

Answer 29

yes thanks

Answer 30

:)