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theta=tan^-1(2tan^2theta)-tan^-1{(1/3)ta...

`theta=tan^-1(2tan^2theta)-tan^-1{(1/3)tantheta}`, if

A

`tantheta=-2`

B

`tantheta=0`

C

`tantheta=1`

D

`tantheta=2`

Text Solution

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The correct Answer is:
A, B, C
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Knowledge Check

  • theta = tan ^(-1)(2 tan ^(2) theta )- tan ^(-1) ((1)/(3) tan theta ) if -

    A
    `tan theta =-2`
    B
    ` tan theta =0`
    C
    ` tan theta=1`
    D
    `tan theta = 2`

  • (1 + tan^(2) theta)/(1 - tan^(2) theta)=

    A
    `cos^(2) theta`
    B
    `cos 2 theta`
    C
    `sec^(2) theta`
    D
    `sec 2 theta`

  • A function f:(0,pi //2) to R is defined as : f(theta )=|(1,tan theta ,1),(-tan theta ,1,tan theta ),(-1,-tan theta ,1)| then the range of f is -

    A
    `(2,oo)`
    B
    `(-oo,-2)`
    C
    `(2,ooo)`
    D
    `(-oo,2)`

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    tan 3theta = tan 2theta + tantheta

    tan3theta + tantheta = 2tan 2theta

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