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tan(45^(@)+ theta )*tan(45^(@)-theta)=...

`tan(45^(@)+ theta )*tan(45^(@)-theta)=`

A

0

B

1

C

`-1`

D

2

Text Solution

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The correct Answer is:
B
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If 3theta is not an odd multiple of pi/2 then prove that tanthetatan(60^(@)+theta) tan(60^(@)-theta)=tan3theta and deduce that tan6^(@) tan42^(@) tan66^(@) tan78^(@)=1 .

If tan theta tan (120^(@)-theta)tan (120^(@)+theta)=1//sqrt(3) , then theta =

Knowledge Check

  • Galileo writes that for angles of projection of a projectile at angles (45^(@)+theta) and (45^(@) -theta) , the horizontal ranges described by the projectile are in the ratio of (if theta le 45^(@))

    A
    `2:1`
    B
    `1:2`
    C
    `1:1`
    D
    `2:3`

  • (1-tan^(2)(45^(@)-theta))/(1+tan^(2)(45^(@)-theta))=

    A
    `sin 2theta`
    B
    `cos 2 theta`
    C
    `tan 2theta`
    D
    `cot2 theta`

  • If tan theta + tan(60^(@)+theta)+tan (120^(@)+theta)=3 , then theta =

    A
    `(4n+1)pi/12`
    B
    `(2n+1)pi//12`
    C
    `n pi+pi//3`
    D
    none

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    (tan(45^(@)+alpha)-tan(45^(@)-alpha))/(cot(45^(@)+alpha)-cot(45^(@)-alpha))=

    ( tan^(3) theta )/( 1+ tan^(2) theta )+ ( cot^(3) theta )/( 1+ cot^(2) theta )=

    If tan theta = -(5)/(12) and theta is not in the fourth quadrant then ( tan (90^(@)+ theta ) - sin(180^(@) - theta ))/( sin (270^(@)- theta ) + " cosec " (360^(@) - theta ))=

    Cos (theta + 45^(@)) + Sin (theta - 45^(@)) =

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