Mathematics Asked on November 9, 2021
I want to simplify the following expression: $frac{{left( {1 + sqrt 3 tan {1^{circ}}} right)left( {1 + sqrt 3 tan {2^{circ}}} right)left( {tan {1^{circ}} + tan {{59}^{circ}}} right)left( {tan {2^{circ}} + tan {{58}^{circ}}} right)}}{{left( {1 + {{tan }^2}{1^{circ}}} right)left( {1 + {{tan }^2}{2^{circ}}} right)}}$
My approach is as follow
$T = left( {1 + sqrt 3 tan {1^{circ}}} right)left( {1 + sqrt 3 tan {2^{circ}}} right)left( {tan {1^{circ}} + tan {{59}^{circ}}} right)left( {tan {2^{circ}} + tan {{58}^{circ}}} right){cos ^2}{1^{circ}}{cos ^2}{2^{circ}}$
$Rightarrow tan {58^{circ}}left( {1 + sqrt 3 tan {2^{circ}}} right) = sqrt 3 – tan {2^{circ}}$
$Rightarrow tan {59^{circ}}left( {1 + sqrt 3 tan {2^0}} right) = sqrt 3 – tan {1}$
$tan {60^{circ}} = frac{{left( {tan {1^{circ}} + tan {{59}^{circ}}} right)}}{{left( {1 + tan {{59}^{circ}}tan {1^{circ}}} right)}} Rightarrow left( {tan {1^{circ}} + tan {{59}^{circ}}} right) = sqrt 3 left( {1 + tan {{59}^{circ}}tan {1^{circ}}} right)$
$tan {60^{circ}} = frac{{left( {tan {2^{circ}} + tan {{58}^{circ}}} right)}}{{left( {1 + tan {{58}^{circ}}tan {2^{circ}}} right)}} Rightarrow left( {tan {2^{circ}} + tan {{58}^{circ}}} right) = sqrt 3 left( {1 + tan {{58}^{circ}}tan {2^{circ}}} right)$
$T = frac{{sqrt 3 – tan {1^{circ}}}}{{tan {{59}^{circ}}}} times frac{{sqrt 3 – tan {2^{circ}}}}{{tan {{58}^{circ}}}} times sqrt 3 left( {1 + tan {{59}^{circ}}tan {1^{circ}}} right) times sqrt 3 left( {1 + tan {{58}^{circ}}tan {2^{circ}}} right){cos ^2}{1^{circ}}{cos ^2}{2^{circ}}$
After this step I am confused
Generalization:
$$dfrac{(1+tan ytan z)(tan z+tan(y-z))}{1+tan^2z} =cdots =dfrac{cos(y-z)sin(z+y-z)cos^2z}{cos ycos zcos zcos(y-z)} =tan y$$
$$text{ if }cos(y-z)cos zne0$$
Here $y=60^circ$
Then $z=1^circ, 2^circ$
Answered by lab bhattacharjee on November 9, 2021
$$ tan 59^{circ} = frac{sqrt{3}-tan 1^{circ}}{1+sqrt{3} tan 1^{circ}} $$ $$ tan 59^{circ}+tan 1^{circ} = frac{sqrt{3}(1+tan^2 1^{circ})}{1+sqrt{3} tan 1^{circ}}, $$so we have $$ (1+sqrt{3} tan 1^{circ})(tan 59^{circ}+tan 1^{circ})=sqrt{3}(1+tan^2 1^{circ}) $$A similar argument works for $tan 2^{circ}$ and $tan 58^{circ}$. So the answer is $3$.
Answered by Namburu Karthik on November 9, 2021
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