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Question: Find the value of \(\tan {9^0} - \tan {27^0} - \tan {63^0} + \tan {81^0}\) \( {\text{A}}{\text...

Find the value of tan⁡90−tan⁡270−tan⁡630+tan⁡810\tan {9^0} - \tan {27^0} - \tan {63^0} + \tan {81^0}
A. 4 B. 3 C. 2 D. 1  {\text{A}}{\text{. 4}} \\\ {\text{B}}{\text{. 3}} \\\ {\text{C}}{\text{. 2}} \\\ {\text{D}}{\text{. 1}} \\\

Explanation

Solution

To solve this type of question you have to do some trigonometric transformation by seeing that the sum of two angles is 90 so these are complementary angles, hence you can transform to make the question easy. And solve further using standard trigonometric results.

Complete step-by-step answer:
We have
tan⁡90−tan⁡270−tan⁡63.0+tan⁡810\tan {9^0} - \tan {27^0} - \tan {63.0} + \tan {81^0}
First take minus common then we get
=tan⁡90+tan⁡810−(tan⁡270+tan⁡630)= \tan {9^0} + \tan {81^0} - \left( {\tan {{27}^0} + \tan {{63}^0}} \right)
Now we will do transformation as
=tan⁡90+tan⁡(900−90)−(tan⁡270+tan⁡(900−270)) [∵810=(900−90)] =tan⁡90+cot⁡90−(tan⁡270+tan⁡270)[∵cot⁡A=tan⁡(900−A)] =sin⁡90cos⁡90+cos⁡90sin⁡90−(sin⁡270cos⁡270+cos⁡270sin⁡270) [∵tan⁡θ=sin⁡θcos⁡θ] =sin⁡290+cos⁡290sin⁡90.cos⁡90−(sin⁡2270+cos⁡2270sin⁡270.cos⁡270)[∵sin⁡2θ+cos⁡2θ=1] =1sin⁡90.cos⁡90−1sin⁡270.cos⁡270  = \tan {9^0} + \tan \left( {{{90}^0} - {9^0}} \right) - \left( {\tan {{27}^0} + \tan \left( {{{90}^0} - {{27}^0}} \right)} \right){\text{ }}\left[ {\because {{81}^0} = \left( {{{90}^0} - {9^0}} \right)} \right] \\\ = \tan {9^0} + \cot {9^0} - \left( {\tan {{27}^0} + \tan {{27}^0}} \right)\left[ {\because \cot A = \tan \left( {{{90}^0} - A} \right)} \right] \\\ = \dfrac{{\sin {9^0}}}{{\cos {9^0}}} + \dfrac{{\cos {9^0}}}{{\sin {9^0}}} - \left( {\dfrac{{\sin {{27}^0}}}{{\cos {{27}^0}}} + \dfrac{{\cos {{27}^0}}}{{\sin {{27}^0}}}} \right){\text{ }}\left[ {\because \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}} \right] \\\ = \dfrac{{{{\sin }^2}{9^0} + {{\cos }^2}{9^0}}}{{\sin {9^0}.\cos {9^0}}} - \left( {\dfrac{{{{\sin }^2}{{27}^0} + {{\cos }^2}{{27}^0}}}{{\sin {{27}^0}.\cos {{27}^0}}}} \right)\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right] \\\ = \dfrac{1}{{\sin {9^0}.\cos {9^0}}} - \dfrac{1}{{\sin {{27}^0}.\cos {{27}^0}}} \\\
Now we know (sin2A= 2sinA.cosA)
This formula is very helpful so we should remember it. Whenever we get multiplication of sine and cosine with the same angle this formula should be applied.
So to apply the above formula multiply and divide by 2 on numerator and denominator.
=22.sin⁡90cos⁡90−22.sin⁡270.cos⁡270= \dfrac{2}{{2.\sin {9^0}\cos {9^0}}} - \dfrac{2}{{2.\sin {{27}^0}.\cos {{27}^0}}}
=2sin⁡180−2sin⁡540=2[sin⁡540−sin⁡180sin⁡180.sin⁡540]= \dfrac{2}{{\sin {{18}^0}}} - \dfrac{2}{{\sin {{54}^0}}} = 2\left[ {\dfrac{{\sin {{54}^0} - \sin {{18}^0}}}{{\sin {{18}^0}.\sin {{54}^0}}}} \right]
Now we will apply the formula (∵sin⁡C−sin⁡D=2cos⁡C+D2.sin⁡C−D2)\left( {\because \sin C - \sin D = 2\cos \dfrac{{C + D}}{2}.\sin \dfrac{{C - D}}{2}} \right)
=2.2[cos⁡540+1802.sin⁡540−1802sin⁡180.sin⁡540]= 2.2\left[ {\dfrac{{\cos \dfrac{{{{54}^0} + {{18}^0}}}{2}.\sin \dfrac{{{{54}^0} - {{18}^0}}}{2}}}{{\sin {{18}^0}.\sin {{54}^0}}}} \right]
=2.2cos⁡360sin⁡180sin⁡180.sin⁡(900−360)=2.2cos⁡360sin⁡180cos⁡360.sin⁡180=4= \dfrac{{2.2\cos {{36}^0}\sin {{18}^0}}}{{\sin {{18}^0}.\sin \left( {{{90}^0} - {{36}^0}} \right)}} = \dfrac{{2.2\cos {{36}^0}\sin {{18}^0}}}{{\cos {{36}^0}.\sin {{18}^0}}} = 4
Hence option A is the correct option.

Note: Whenever we get this type of question the key concept of solving is you have to start a solution from simple trigonometric transformation and you have to use many formulas like addition to multiplication half angle formula and many more so we should remember all these to solve this type of question.