The value of log _{10} tan 1^{circ} + log _{1 | vedclass

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Question

(A) Using the property $\log a + \log b = \log(ab)$,the expression becomes: $\log _{10} (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \dots \cdot 	an 89

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Using the property $\log a + \log b = \log(ab)$,the expression becomes: $\log _{10} (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \dots \cdot 	an 89^{\circ})$ We know that $	an 	heta \cdot 	an(90^{\circ} - 	heta) = 	an 	heta \cdot \cot 	heta = 1$. Pairing the terms: $(	an 1^{\circ} \cdot 	an 89^{\circ}) \cdot (	an 2^{\circ} \cdot 	an 88^{\circ}) \cdot \dots \cdot (	an 44^{\circ} \cdot 	an 46^{\circ}) \cdot 	an 45^{\circ}$ $= (1) \cdot (1) \cdot \dots \cdot (1) \cdot 1 = 1$ Therefore,$\log _{10} (1) = 0$.

The value of $\log _{10} \tan 1^{\circ} + \log _{10} \tan 2^{\circ} + \dots + \log _{10} \tan 89^{\circ}$ is equal to :-

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