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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

The value of tan 63 ^ o - cot 63 ^ o is equal to

The value of tan63o−cot63otan {63}^{o} - cot {63}^{o} is equal to

Options

  1. A25+110−25\dfrac{2}{\sqrt{5} + 1}\sqrt{10 - 2\sqrt{5}}
  2. B25+110+25\dfrac{2}{\sqrt{5} + 1}\sqrt{10 + 2\sqrt{5}}
  3. C5−1410−25\dfrac{\sqrt{5} - 1}{4}\sqrt{10 - 2\sqrt{5}}
  4. D5−1410+25\dfrac{\sqrt{5} - 1}{4}\sqrt{10 + 2\sqrt{5}}

Correct answer

A. 25+110−25\dfrac{2}{\sqrt{5} + 1}\sqrt{10 - 2\sqrt{5}}

Step-by-step solution

sin63ocos63o−cos63osin63o=sin263o−cos⁡263ocos63osin63o\dfrac{sin {63}^{o}}{cos {63}^{o}} - \dfrac{cos {63}^{o}}{sin {63}^{o}} = \dfrac{sin^{2}{63}^{o} - \cos^{2} {63}^{o}}{cos {63}^{o}sin {63}^{o}}
=−cos126o12sin126o=−2cot126o= - \dfrac{cos {126}^{o}}{\dfrac{1}{2}sin {126}^{o}} = - 2cot {126}^{o}
=−2cot⁡(π2+36∘)=2tan⁡36∘=2sin⁡36∘cos⁡36∘= - 2\cot \left(\dfrac{\pi}{2} + 3{6}^{∘}\right) = 2\tan 3{6}^{∘} = \dfrac{2\sin 3{6}^{∘}}{\cos 3{6}^{∘}}

=2(5+1)10−25= \dfrac{2}{\left(\sqrt{5} + 1\right)}\sqrt{10 - 2\sqrt{5}}
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