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If tan 20^@ = lambda, then (tan160^@ - t...

If `tan 20^@ = lambda`, then `(tan160^@ - tan 110^@)/(1+ (tan 160^@)(tan110^@)` =

A

`(1+lambda^2)/(2lambda)`

B

`(1+lambda^2)/lambda`

C

`(1-lambda^2)/lambda`

D

`(1-lambda^2)/(2lambda)`

Text Solution

Verified by Experts

The correct Answer is:
D
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If tan20^(@) = lambda then show that (tan 160^(@) -tan 110^(@))/(1+tan160^(@).tan110^(@)) = (1-lambda^2)/(2lambda) .

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Knowledge Check

  • If tan 20^@ =p , " then " (tan 160^@ - tan 110^(@))/(1+ tan 160^(@) tan 110^(@))=

    A
    `(2p)/(1+p^(2))`
    B
    `(1-p^(2))/(2p)`
    C
    `(1-p^(2))/(1+p^(2))`
    D
    `(2p)/(1-p^(2))`

  • If tan 40^(@)= lambda , then (tan 140^(@)-tan 130^(@))/(1+ tan 140^(@) tan 130^(@))=

    A
    `(1-lambda^(2))/(lambda)`
    B
    `(1+lambda^(2))/(lambda)`
    C
    `(1+lambda^(2))/(2lambda)`
    D
    `(1-lambda^(2))/(2lambda)`

  • If tan 20^(@) = lambda , " then " (tan 250^@ + tan 340^(@))/(tan 200^(@) - tan 110^(@))=

    A
    `(1+ lambda^(2))/(1-lambda^(2))`
    B
    `(1- lambda^(2))/(1+lambda^(2))`
    C
    `(1+lambda^(2))/(2lambda)`
    D
    `(1-lambda^(2))/(2lambda)`

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