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    The value of tan⁡85°−tan⁡25°(1+tan85°tan⁡25°)\frac{tan⁡85°-tan⁡25°}{(1+tan85°tan⁡25° )}(1+tan85°tan⁡25°)tan⁡85°−tan⁡25°​​  is:
    Question

    The value of tan85°tan25°(1+tan85°tan25°)\frac{tan⁡85°-tan⁡25°}{(1+tan85°tan⁡25° )}​  is:

    A.

    1/√3

    B.

    1

    C.

    √3

    D.

    0

    Correct option is C

    Given:

    tan85tan251+tan85tan25\frac{\tan 85^\circ - \tan 25^\circ}{1 + \tan 85^\circ \cdot \tan 25^\circ}​​

    Formula Used:

    tanAtanB1+tanAtanB=tan(AB)\frac{\tan A - \tan B}{1 + \tan A \cdot \tan B} = \tan (A - B)​​

    Solution:

    tanAtanB1+tanAtanB=tan(AB)\frac{\tan A - \tan B}{1 + \tan A \cdot \tan B} = \tan (A - B)​​

    tan85tan251+tan85tan25=tan(8525)\frac{\tan 85^\circ - \tan 25^\circ}{1 + \tan 85^\circ \cdot \tan 25^\circ} = \tan (85^\circ - 25^\circ)​​

    tan60=3\tan 60^\circ = \sqrt{3}​​

    Thus, the value of the given expression =3\sqrt{3}​​

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