The value of \tan 9^\circ-\tan 27^\circ-\tan 63^\circ+\tan 81^\circ is equal to

NIMCET 2021 Question Paper with solutions

Question : 45 of 120

Marks: +1, -0

The value of

\tan 9°-\tan 27°-\tan 63°+\tan 81°

is equal to

Solution:

\tan 9°-\tan 27°-\tan 63°+\tan 81°

⇒\tan 9°+\tan 81°-(\tan 27°+\tan 63°)

\∴\tan A+\tan B={\sin (A+B)}/{\cos A \cos B}

\⇒{\sin (9+81)}/{\cos 9 \cos 81}-{\sin (27+63)}/{\cos 27 \cos 63}

\⇒{2 \sin 90}/{2 \cos 9 \sin 9} {2 \sin 90}/{2 \cos 27 \sin 27}

[\∴\sin \θ=\cos (90-\θ)]

\⇒ {2}/{\sin 18°}-{2}/{\sin 54°}

[\∴2 \sin \θ\cos \θ=\sin 2 \θ]

⇒{2}/{{\√{5}-1}/{4}}-{2}/{{\√{5}+1}{4}}

\sin 18°={\√{5}-1}/{4}\ \&\ \sin 54°={\√{5}+1}/{4}

\⇒8[{1}/{\√{5}-1}-{1}/{\√{5}+1}]

\⇒8[{2}/{4}]

=4

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