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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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