Sham is trying to solve the expression:
$$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$.
The correct answer would be?

$$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$.

=$$\log \tan 1^\circ + \log \tan 89^\circ + \log \tan 2^\circ + \log \tan 88^\circ ........ + \log \tan 45^\circ$$.

=$$\log\ \left(\tan\ 1^0\cdot\tan\ 89^0\right)\times\log\ \left(\tan\ 2^0\cdot\tan\ 88^0\right)\ ...........................\log\ \left(\tan\ 45^0\right)$$

tan $$45^0$$ = 1

$$\log\ \left(\tan\ 45^0\right)\ =\ 0$$

$$\therefore$$ $$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$ = 0