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
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