SarkariPYQ

SSC CGL 2022 PYQ — Question 70

SSC CGL - 01 Dec 2022 - Shift 1 • Quantitative Aptitude • Q70

Back to paper

Simplify cos⁡45∘sec⁡30∘+cosec⁡30∘\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}

Show answer & explanation

Correct answer: C

Explanation

The correct answer is Option 3. i.e  32−68\frac{3 \sqrt{2}-\sqrt{6}}{8}.

cos⁡45∘sec⁡30∘+cosec⁡30∘\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}

=1223+2=122+233=32(2+23)=322+26=3(26−22)((26)+22)(26−22)=23(6−2)(26)2−(22)2=23(6−2)24−8=23(6−2)16=18−68=32−68\begin{aligned} & =\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}} \\ & =\frac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\frac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}} \\ & =\frac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{((2 \sqrt{6})+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})} \\ & =\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^2-(2 \sqrt{2})^2}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16} \\ & =\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\end{aligned}