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Question

Find the value of $$\tan  9^{\circ}-\tan  27^{\circ}-\tan  63^{\circ}+\tan  81^{\circ}.$$ 


A
4
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B
3
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C
2
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D
1
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Solution

The correct option is D 4
$$\tan { 9° } -\tan { 27° } -\tan { 63° } +\tan { 81° } $$

$$=\tan { 9° } +\tan { 81° } -\left( \tan { 27° } +\tan { 63° }  \right) $$

$$=\tan { 9° } +\tan { (90°-9°) } -\left( \tan { 27° } +\tan { (90°-27°) }  \right) $$

$$=\tan { 9° } +\cot { 9° } -\left( \tan { 27° } +\tan { 27° }  \right) $$     $$[\because \cot A = \tan(90˚-A)]$$

$$=\dfrac{\sin 9°}{\cos 9°}+\dfrac{\cos 9°}{\sin 9°}-\left(\dfrac{\sin 27°}{\cos 27°} +\dfrac{\cos 27°}{\sin 27°}\right)$$

$$=\cfrac { \sin ^{ 2 }{ 9° } +\cos ^{ 2 }{ 9° }  }{ \sin { 9° } \cos { 9° }  } -\left( \cfrac { \sin ^{ 2 }{ 27° } +\cos ^{ 2 }{ 27° }  }{ \sin { 27° } \cos { 27° }  }  \right) $$

$$=\cfrac { 1 }{ \sin { 9° } \cos { 9° }  } -\cfrac { 1 }{ \sin { 27° } \cos { 27° }  } =\cfrac { 2 }{ \sin { 18° }  } -\cfrac { 2 }{ \sin { 54° }  } $$    $$[\because \sin 2x =2\sin x \cos x]$$

$$=2\left[\dfrac{\sin { 54° } -\sin { 18° }}{\sin { 18° } \sin { 54° } }\right]=\dfrac{2.2\cos { 36° } \sin { 18° }  }{\sin { 18° } \sin { (90°-36°) }}$$      $$\left[\because \sin C - \sin D = 2 cos \dfrac {C+D}{2} \sin \dfrac {C-D}{2}\right]$$

$$=\cfrac { 2.2\cos { 36° } \sin { 18° }  }{ \cos { 36° } \sin { 18° }  } =4$$

Mathematics

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