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Question

Question: \[\mathbf{\tan}\mathbf{1}\mathbf{00}\mathbf{{^\circ}}\mathbf{+}\mathbf{\tan}\mathbf{1}\mathbf{25}\ma...

tan⁡100∘+tan⁡125∘+tan⁡100∘tan⁡125∘=\mathbf{\tan}\mathbf{1}\mathbf{00}\mathbf{{^\circ}}\mathbf{+}\mathbf{\tan}\mathbf{1}\mathbf{25}\mathbf{{^\circ}}\mathbf{+}\mathbf{\tan}\mathbf{1}\mathbf{00}\mathbf{{^\circ}}\mathbf{\tan}\mathbf{1}\mathbf{25}\mathbf{{^\circ}}\mathbf{=}

A

0

B

1/2

C

– 1

D

1

Answer

1

Explanation

Solution

tan⁡(100o+125o)=tan⁡100o+tan⁡125o1−tan⁡100otan⁡125o\tan(100^{o} + 125^{o}) = \frac{\tan 100^{o} + \tan 125^{o}}{1 - \tan 100^{o}\tan 125^{o}}

∴\therefore tan⁡225o=tan⁡100o+tan⁡125o1−tan⁡100otan⁡125o\tan 225^{o} = \frac{\tan 100^{o} + \tan 125^{o}}{1 - \tan 100^{o}\tan 125^{o}}

i.e., 1=tan⁡100o+tan⁡125o1−tan⁡100otan⁡125o1 = \frac{\tan 100^{o} + \tan 125^{o}}{1 - \tan 100^{o}\tan 125^{o}}

i.e.,tan⁡100o+tan⁡125o+tan⁡100otan⁡125o=1.\tan 100^{o} + \tan 125^{o} + \tan 100^{o}\tan 125^{o} = 1.