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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If 4cos 36 ^ o + (7(1)/(2) ∘) = n _ 1 + n _ 2 + n _ 3 + n _ 4 + n _...

If 4cos36o+cot⁡(712∘)=n1+n2+n3+n4+n5+n64cos {36}^{o} + \cot \left(7\dfrac{1}{2} ∘\right) = \sqrt{{n}_{1}} + \sqrt{{n}_{2}} + \sqrt{{n}_{3}}+ \sqrt{{n}_{4}} + \sqrt{{n}_{5}} + \sqrt{{n}_{6}}, then the value of \left(\dfrac{\mathop{∑}_{i = 1}^{6}{n}_{i}^{2}}{10}\right) is equal to

Correct answer

9.1

Step-by-step solution

4cos36o=4(5+14)=5+14cos {36}^{o} = 4\left(\dfrac{\sqrt{5} + 1}{4}\right) = \sqrt{5} + \sqrt{1}
cot⁡712o=1+cos15°sin15°=1+(3+122)(3−122)=22+3+13−1\cot 7\dfrac{1}{2}o = \dfrac{1 + cos {15}^{°}}{sin {15}^{°}} = \dfrac{1 + \left(\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\right)}{\left(\dfrac{\sqrt{3} - 1}{2\sqrt{2}}\right)} = \dfrac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}
=(22+3+1)(3+1)(3−1)(3+1)=26+22+3+3+3+12= \dfrac{\left(2\sqrt{2} + \sqrt{3} + 1\right)\left(\sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right)\left(\sqrt{3} + 1\right)} = \dfrac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{2}
=26+23+22+42=6+3+2+2=6+4+3+2= \dfrac{2\sqrt{6} + 2\sqrt{3} + 2\sqrt{2} + 4}{2} = \sqrt{6} + \sqrt{3} + \sqrt{2} + 2 = \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}
⇒4cos36o+cot⁡(712o)=5+1+6+4+3+2⇒ 4cos {36}^{o} + \cot \left(7{\dfrac{1}{2}}^{o}\right) = \sqrt{5} + \sqrt{1} + \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}
=1+2+3+4+5+6= \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{5} + \sqrt{6}
=n1+n2+n3+n4+n5+n6= \sqrt{{n}_{1}} + \sqrt{{n}_{2}} + \sqrt{{n}_{3}} + \sqrt{{n}_{4}} + \sqrt{{n}_{5}} + \sqrt{{n}_{6}}
⇒∑i=16ni2=12+22+33+42+52+62⇒ \mathop{∑}_{i = 1}^{6}{n}_{i}^{2} = {1}^{2} + {2}^{2} + {3}^{3} + {4}^{2} + {5}^{2} + {6}^{2}
=6×7×136=91= \dfrac{6 \times 7 \times 13}{6} = 91
⇒∑i=16ni210=9110=9.1⇒ \dfrac{\mathop{∑}_{i = 1}^{6}{n}_{i}^{2}}{10} = \dfrac{91}{10} = 9 . 1
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