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  1. The value of tan
    13π
    		will be
    12

    1. √3 + 1
      √3 - 1
    2. √3 - 1
      √3 + 1
    3. √3
      √3 + 1
    4. 1
      √3 - 1
Correct Option: B

tan
13π
12

= tanπ +
π
12

= tan
π
12

[∵ tan(π + θ) = tanθ]
= tan
π
 -
π
34

∵
1
 =
1
 -
1
1234

= tan
π
 - tan
π
33
1 + tan
π
tan
π
34

∵ tan (A – B)
=
tanA - tanB
1 + tanA tanB

=
√3 - 1
1 + √3

=
√3 - 1
√3 + 1


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