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SSC CGL 2023 PYQ — Question 65

SSC CGL - 19 July 2023 - Shift 2 • Quantitative Aptitude • Q65

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Using cot(A - B) =cotAcotB+1cotB−cotA=\frac{cotAcotB+1}{cotB-cotA}, find the value of cot15°.

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Correct answer: A

Explanation

The correct answer is Option 1 i.e. 2+32+\sqrt{3}.

cot(A - B) =cotAcotB+1cotB−cotA=\frac{cotAcotB+1}{cotB-cotA} 

⇒ cot 15° = cot (45° - 30°)

⇒ cot (45° - 30°) =cot45°cot30°+1cot30°−cot45°=\frac{cot45°cot30°+1}{cot30°-cot45°}

⇒ cot 15° =(1)(3)+1(3)−(1)=\frac{(1)(\sqrt3)+1}{(\sqrt3)-(1)}

⇒ cot 15° =(3)+1(3)−(1)×(3)+1(3)+(1)={\frac{(\sqrt3)+1}{(\sqrt3)-(1)}} \times {\frac{(\sqrt3)+1}{(\sqrt3)+(1)}}

⇒ cot 15° =(3+1)2(3)2−12=\frac {(\sqrt3 + 1)^2}{(\sqrt3)^2 - 1^2}

⇒ cot 15° =3+1+233−1=\frac {3 + 1 + 2\sqrt3}{3 - 1}

⇒ cot 15° =4+232=\frac {4 + 2\sqrt3}{2}

⇒ cot 15° =2(2+3)2=\frac {2(2 + \sqrt3)}{{2}}

⇒ cot 15° =(2+3)={(2 + \sqrt3)}