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Solution

Using property: $\log a + \log b = \log(ab)$ So expression becomes: $e^{\log 10 \left(\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ\right)}$ Using identity: $\tan \theta \tan (90^\circ-\theta) = 1$ All terms cancel pairwise: $\tan 1^\circ \tan 89^\circ \cdot \tan 2^\circ \tan 88^\circ \cdots = 1$ Thus exponent becomes $\log 10 (1)=0$ So value $= e^0 = 1$