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UPSC NDA Math Model Paper 1

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Question : 41 of 120

Marks: +1, -0

\tan 36^{\circ}+\tan 9^{\circ}+\tan 36^{\circ}\cdot\tan 9^{\circ}=?

-1
1
2
-2

Solution:

To find the value of

\tan 36^{\circ}+\tan 9^{\circ}+\tan 36^{\circ}\cdot\tan 9^{\circ},

consider

A=36^{\circ}

and

B=9^{\circ}

We know that,

\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\cdot\tan B}

\Rightarrow \tan(36^{\circ}+9^{\circ})=\frac{\tan 36^{\circ}+\tan 9^{\circ}}{1-\tan 36^{\circ}\cdot\tan 9^{\circ}}

\Rightarrow\tan(45^{\circ})=\frac{\tan 36^{\circ}+\tan 9^{\circ}}{1-\tan 36^{\circ}\cdot\tan 9^{\circ}}

\Rightarrow 1=\frac{\tan 36^{\circ}+\tan 9^{\circ}}{1-\tan 36^{\circ}\cdot\tan 9^{\circ}}

\Rightarrow 1-\tan 36^{\circ}\cdot\tan 9^{\circ}=\tan 36^{\circ}+\tan 9^{\circ}

\Rightarrow \tan 36^{\circ}+\tan 9^{\circ}+\tan 36^{\circ}\cdot\tan 9^{\circ}=1

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