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Question: The value of \(\tan {81^0} - \tan {63^0} - \tan {27^0} + \tan {9^0}\)is \( \left( A \right).{\...

The value of tan⁡810−tan⁡630−tan⁡270+tan⁡90\tan {81^0} - \tan {63^0} - \tan {27^0} + \tan {9^0}is
(A). 0 (B). 2 (C). 3 (D). 4  \left( A \right).{\text{ 0}} \\\ \left( B \right).{\text{ 2}} \\\ \left( C \right).{\text{ 3}} \\\ \left( D \right).{\text{ 4}} \\\

Explanation

Solution

Hint: Try to write tan⁡θ\tan \theta in terms of cot⁡θ\cot \theta , then use trigonometric formulas.

Given:tan⁡810−tan⁡630−tan⁡270+tan⁡90\tan {81^0} - \tan {63^0} - \tan {27^0} + \tan {9^0}
Rewriting above as:
⇒(tan⁡90+tan⁡810)−(tan⁡270+tan⁡630) ⇒(tan⁡90+tan⁡(900−90))−(tan⁡270+tan⁡(900−270)) …(A)  \Rightarrow \left( {\tan {9^0} + \tan {{81}^0}} \right) - \left( {\tan {{27}^0} + \tan {{63}^0}} \right) \\\ \Rightarrow \left( {\tan {9^0} + \tan \left( {{{90}^0} - {9^0}} \right)} \right) - \left( {\tan {{27}^0} + \tan \left( {{{90}^0} - {{27}^0}} \right)} \right){\text{ }} \ldots \left( A \right) \\\
We know that, tan⁡(900−θ)=cot⁡θ=cos⁡θsin⁡θ …(1)\tan \left( {{{90}^0} - \theta } \right) = \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}{\text{ }} \ldots \left( 1 \right)
Putting the value of tan⁡(900−θ)\tan \left( {{{90}^0} - \theta } \right)from (1)\left( 1 \right)in (A)\left( A \right), we get
⇒tan⁡90+cot⁡90−tan⁡270−cot⁡270\Rightarrow \tan {9^0} + \cot {9^0} - \tan {27^0} - \cot {27^0}
Now, since tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}

∴(sin⁡90cos⁡90+cos⁡90sin⁡90)−(sin⁡270cos⁡270+cos⁡270sin⁡270) ⇒(sin⁡290+cos⁡290sin⁡90cos⁡90)−(sin⁡2270+cos⁡2270sin⁡270cos⁡270)  \therefore \left( {\dfrac{{\sin {9^0}}}{{\cos {9^0}}} + \dfrac{{\cos {9^0}}}{{\sin {9^0}}}} \right) - \left( {\dfrac{{\sin {{27}^0}}}{{\cos {{27}^0}}} + \dfrac{{\cos {{27}^0}}}{{\sin {{27}^0}}}} \right) \\\ \Rightarrow \left( {\dfrac{{{{\sin }^2}{9^0} + {{\cos }^2}{9^0}}}{{\sin {9^0}\cos {9^0}}}} \right) - \left( {\dfrac{{{{\sin }^2}{{27}^0} + {{\cos }^2}{{27}^0}}}{{\sin {{27}^0}\cos {{27}^0}}}} \right) \\\

Using the property sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 in above equation, we get
⇒(1sin⁡90cos⁡90)−(1sin⁡270cos⁡270)\Rightarrow \left( {\dfrac{1}{{\sin {9^0}\cos {9^0}}}} \right) - \left( {\dfrac{1}{{\sin {{27}^0}\cos {{27}^0}}}} \right)
Also, we know that sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \dfrac{{\sin 2x}}{2}

∴(2sin⁡180)−(2sin⁡540) ⇒2(sin⁡540−sin⁡180sin⁡180sin⁡540)  \therefore \left( {\dfrac{2}{{\sin {{18}^0}}}} \right) - \left( {\dfrac{2}{{\sin {{54}^0}}}} \right) \\\ \Rightarrow 2\left( {\dfrac{{\sin {{54}^0} - \sin {{18}^0}}}{{\sin {{18}^0}\sin {{54}^0}}}} \right) \\\

Now, by using sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) in above equation, we get
⇒2(2cos⁡(54+182)sin⁡(54−182)sin⁡180sin⁡(900−360))\Rightarrow 2\left( {\dfrac{{2\cos \left( {\dfrac{{54 + 18}}{2}} \right)\sin \left( {\dfrac{{54 - 18}}{2}} \right)}}{{\sin {{18}^0}\sin \left( {{{90}^0} - {{36}^0}} \right)}}} \right)
using sin⁡(900−θ)=cos⁡θ\sin \left( {{{90}^0} - \theta } \right) = \cos \theta in the above equation, we get

⇒2(2cos⁡(722)sin⁡(362)sin⁡180cos⁡360) ⇒2(2cos⁡(360)sin⁡(180)sin⁡180cos⁡360) ⇒4(sin⁡180cos⁡360sin⁡180cos⁡360) ⇒4  \Rightarrow 2\left( {\dfrac{{2\cos \left( {\dfrac{{72}}{2}} \right)\sin \left( {\dfrac{{36}}{2}} \right)}}{{\sin {{18}^0}\cos {{36}^0}}}} \right) \\\ \Rightarrow 2\left( {\dfrac{{2\cos \left( {{{36}^0}} \right)\sin \left( {{{18}^0}} \right)}}{{\sin {{18}^0}\cos {{36}^0}}}} \right) \\\ \Rightarrow 4\left( {\dfrac{{\sin {{18}^0}\cos {{36}^0}}}{{\sin {{18}^0}\cos {{36}^0}}}} \right) \\\ \Rightarrow 4 \\\

∴\therefore the correct option is (D)\left( D \right).

Note: Whenever there are integer angles inside trigonometric functions whose values are not known to us, always try to convert them by using 900−θ{90^0} - \theta in order to make calculation easier.