We solve the given expression step by step. Given expression: \[ e^{\log_{10} \tan 1^\circ + \log_{10} \tan 2^\circ + \log_{10} \tan 3^\circ + \dots + \log_{10} \tan 89^\circ} \] Step 1: Using Logarithmic Property We use the logarithmic identity: \[ \log a + \log b = \log (a \cdot b) \] Applying this to our given expression: \[ \log_{10} \tan 1^\circ + \log_{10} \tan 2^\circ + \dots + \log_{10} \tan 89^\circ = \log_{10} (\tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 89^\circ) \] Thus, the given expression simplifies to: \[ e^{\log_{10} (\tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 89^\circ)} \] Using the logarithmic property \( e^{\log_{b} x} = x \), we get: \[ \tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 89^\circ \] Step 2: Evaluating the Product We use the identity: \[ \tan x \tan (90^\circ - x) = 1 \] Pairing terms: \[ (\tan 1^\circ \tan 89^\circ) \cdot (\tan 2^\circ \tan 88^\circ) \cdot \dots \cdot (\tan 44^\circ \tan 46^\circ) \cdot \tan 45^\circ \] Since each pair simplifies to 1 and \(\tan 45^\circ = 1\), the entire product simplifies to: \[ 1 \times 1 \times 1 \times \dots \times 1 = 1 \] Final Answer: \[ \boxed{1} \]