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Question: Prove that \(\tan 82{\dfrac{1}{2}^ \circ } = (\sqrt 3 + \sqrt 2 )(\sqrt 2 + 1)\)...

Prove that tan⁡8212∘=(3+2)(2+1)\tan 82{\dfrac{1}{2}^ \circ } = (\sqrt 3 + \sqrt 2 )(\sqrt 2 + 1)

Explanation

Solution

We will be making use of few trigonometric and algebraic identities to solve this problem.
Here in this question some trigonometric and algebraic identities must be known which are as mentioned below: -
1. tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}
2. (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
3. Roots of quadratic equation can find out by sridharacharya formula
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step-by-step solution:
As we have to prove tangent trigonometric function the possible angles that
can be used are tan⁡30∘=13\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} and tan⁡60∘=3\tan {60^ \circ } = \sqrt 3 so we will
use tan⁡30∘\tan {30^ \circ } so that we can get half angle relation.
tan⁡30=13\tan 30 = \dfrac{1}{{\sqrt 3 }} ......................equation (1)
(With the help of equation (1) we will find out tan⁡15∘\tan {15^ \circ } which we can also write as tan⁡712∘\tan 7{\dfrac{1}{2}^ \circ } ⇒2tan⁡15∘1−tan⁡215∘=13 \Rightarrow \dfrac{{2\tan {{15}^ \circ }}}{{1 - {{\tan }^2}{{15}^ \circ }}} = \dfrac{1}{{\sqrt 3 }} (Applying identity tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}} where A=15∘A = {15^ \circ } )
Let’s take tan⁡15∘=a\tan {15^ \circ } = a so that we can make quadratic equation and find value of tan⁡15∘\tan {15^ \circ } ⇒2a1−a2=13 \Rightarrow \dfrac{{2a}}{{1 - {a^2}}} = \dfrac{1}{{\sqrt 3 }}
Now cross multiplying
⇒23a=1−a2\Rightarrow 2\sqrt 3 a = 1 - {a^2}
(Shifting terms so that we can form quadratic equation)
⇒a2+23a−1=0\Rightarrow {a^2} + 2\sqrt 3 a - 1 = 0
Now applying sridharacharya formula to find roots of the equation x=−b±b2\-4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} \- 4ac} }}{{2a}} where a=1,b=23,c=−1a = 1,b = 2\sqrt 3 ,c = - 1
⇒x=−23±(23)2−4×1×−12×1\Rightarrow x = \dfrac{{ - 2\sqrt 3 \pm \sqrt {{{(2\sqrt 3 )}^2} - 4 \times 1 \times - 1} }}{{2 \times 1}}
⇒x=−23±4×3+42\Rightarrow x = \dfrac{{ - 2\sqrt 3 \pm \sqrt {4 \times 3 + 4} }}{2}
⇒x=−23±162\Rightarrow x = \dfrac{{ - 2\sqrt 3 \pm \sqrt {16} }}{2}
⇒x=−23±42\Rightarrow x = \dfrac{{ - 2\sqrt 3 \pm 4}}{2}
Two roots will be there x1=−23+42,x2=−23−42{x_1} = \dfrac{{ - 2\sqrt 3 + 4}}{2},{x_2} = \dfrac{{ - 2\sqrt 3 - 4}}{2}
⇒x1=2−3,x2=−2−3\Rightarrow {x_1} = 2 - \sqrt 3 ,{x_2} = - 2 - \sqrt 3
We will take positive root because tan⁡15∘>0\tan {15^ \circ } > 0
∴tan⁡15∘=2−3\therefore \tan {15^ \circ } = 2 - \sqrt 3
Now applying tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}identity again where A= tan⁡712∘\tan 7{\dfrac{1}{2}^ \circ }
⇒2tan⁡712∘1−tan⁡2712∘=2−3\Rightarrow \dfrac{{2\tan 7{{\dfrac{1}{2}}^ \circ }}}{{1 - {{\tan }^2}7{{\dfrac{1}{2}}^ \circ }}} = 2 - \sqrt 3
Let’s tan⁡712∘=b\tan 7{\dfrac{1}{2}^ \circ } = b so that we can make quadratic equation and can solve with
less complexity. ⇒2b1−b2=2−3\Rightarrow \dfrac{{2b}}{{1 - {b^2}}} = 2 - \sqrt 3
Now cross multiplying
⇒2b=(2−3)(1−b2)\Rightarrow 2b = (2 - \sqrt 3 )(1 - {b^2})
⇒2b=2−3−b2(2−3)\Rightarrow 2b = 2 - \sqrt 3 - {b^2}(2 - \sqrt 3 )
⇒b2(2−3)+2b−(2−3)=0\Rightarrow {b^2}(2 - \sqrt 3 ) + 2b - (2 - \sqrt 3 ) = 0
Now applying sridharacharya formula to find roots of the equation x=−b±b2\-4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} \- 4ac} }}{{2a}} where a=2−3,b=2,c=−2+3a = 2 - \sqrt 3 ,b = 2,c = - 2 + \sqrt 3
⇒x=−2±4+4(2−3)22×(2−3)\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 4{{(2 - \sqrt 3 )}^2}} }}{{2 \times (2 - \sqrt 3 )}} (Taking
positive root because tan⁡712∘>0\tan 7{\dfrac{1}{2}^ \circ } > 0 )
⇒−1+8−43(2−3)\Rightarrow \dfrac{{ - 1 + \sqrt {8 - 4\sqrt 3 } }}{{(2 - \sqrt 3 )}} (Simplifying further)
∴tan⁡712∘=6−2−12−3\therefore \tan 7{\dfrac{1}{2}^ \circ } = \dfrac{{\sqrt 6 - \sqrt 2 - 1}}{{2 - \sqrt 3 }}
tan⁡8212∘=cot⁡712∘=(2−36−2−1)\tan 82{\dfrac{1}{2}^ \circ } = \cot 7{\dfrac{1}{2}^ \circ } = (\dfrac{{2 - \sqrt 3 }}{{\sqrt 6 - \sqrt 2 - 1}})
⇒(2−3)(6−(2+1))(6+2+1)(6+2+1)\Rightarrow \dfrac{{(2 - \sqrt 3 )}}{{(\sqrt 6 - (\sqrt 2 + 1))}}\dfrac{{(\sqrt 6 + \sqrt 2 + 1)}}{{(\sqrt6 + \sqrt 2 + 1)}} (Rationalising the denominator term)⇒26−32+22−6+2−3(6−(2+1))2 \Rightarrow \dfrac{{2\sqrt 6 -3\sqrt 2 + 2\sqrt 2 - \sqrt 6 + 2 - \sqrt 3 }}{{{{(\sqrt 6 - (\sqrt 2 + 1))}^2}}} (Multiplying numerator terms)
⇒6−2+2−33−22\Rightarrow \dfrac{{\sqrt 6 - \sqrt 2 + 2 - \sqrt 3 }}{{3 - 2\sqrt 2 }}
⇒(2−1)(3+2)(2−1)2\Rightarrow \dfrac{{(\sqrt 2 - 1)(\sqrt 3 + \sqrt 2 )}}{{{{(\sqrt 2 - 1)}^2}}}
⇒(3+2)(2−1)\Rightarrow \dfrac{{(\sqrt 3 + \sqrt 2 )}}{{(\sqrt 2 - 1)}}
(Rationalising denominator term) ⇒(3+2)(2−1)×(2+1)(2+1)=(3+2)(2+1) \Rightarrow \dfrac{{(\sqrt 3 + \sqrt 2 )}}{{(\sqrt 2 - 1)}} \times\dfrac{{(\sqrt 2 + 1)}}{{(\sqrt 2 + 1)}} = (\sqrt 3 + \sqrt 2 )(\sqrt 2 + 1)
∴tan⁡8212∘=(3+2)(2+1)\therefore \tan 82{\dfrac{1}{2}^ \circ } = (\sqrt 3 + \sqrt 2 )(\sqrt 2 + 1)
Hence it is proved that tan⁡8212∘=(3+2)(2+1)\tan 82{\dfrac{1}{2}^ \circ } = (\sqrt 3 + \sqrt 2 )(\sqrt 2 + 1)

Note: -Some students may find difficulty in converting tan⁡8212∘=cot⁡712∘\tan 82{\dfrac{1}{2}^ \circ } = \cot7{\dfrac{1}{2}^ \circ } so below explanation of this conversion is mentioned so that students can avoid this mistake. tan⁡8212∘=tan⁡(90∘−712∘)=cot⁡712∘\tan 82{\dfrac{1}{2}^ \circ } = \tan ({90^ \circ } - 7{\dfrac{1}{2}^ \circ }) = \cot7{\dfrac{1}{2}^ \circ } (As cot is positive in the first quadrant)
All students must know signs of different trigonometric functions in all four quadrants so that
conversion of angle becomes easy with that knowledge. Below all signs quadrant wise are
mentioned: -
First quadrant = All trigonometric functions are positive (sine, cosine, tan, sec, cosec, cot)
Second quadrant=Positive (sine, cosec).Negative (cosine, tan, sec, cot)
Third quadrant= Positive (tan, cot). Negative (sine, cosine, sec, cosec)
Fourth quadrant= Positive (cosine, sec). Negative (sine, tan, cot, cosec)