Question : Simplify $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
Option 1: $\frac{3 \sqrt{2}+\sqrt{6}}{8}$
Option 2: $\frac{\sqrt{3}}{2 \sqrt{2}-2 \sqrt{6}}$
Option 3: $\frac{3 \sqrt{2}-\sqrt{6}}{8}$
Option 4: $\frac{\sqrt{3}}{2 \sqrt{6}-2 \sqrt{2}}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\frac{3 \sqrt{2}-\sqrt{6}}{8}$
Solution : $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$ = $\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$ = $\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}$ = $\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}\times\frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{2}(\sqrt{3}-1)}$ = $\frac{3 \sqrt{2}-\sqrt{6}}{8}$ Hence, the correct answer is $\frac{3 \sqrt{2}-\sqrt{6}}{8}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : Using $\operatorname{cosec}(\alpha+\beta)=\frac{\sec \alpha \times \sec \beta \times \operatorname{cosec} \alpha \times \operatorname{cosec} \beta}{\sec \alpha \times \operatorname{cosec} \beta+\operatorname{cosec} \alpha \times \sec \beta}$, find the value of
Question : If $x\cos^{2}30^{\circ}\cdot \sin60^{\circ}=\frac{\tan^{2}45^{\circ}\cdot \sec60^{\circ}}{\operatorname{cosec}60^{\circ}}$, then the value of $x$ is:
Question : The value of $\operatorname{cosec} 30^{\circ}-\cos 60^{\circ}$ is:
Question : What is the value of $\frac{\cos 50^{\circ}}{\sin 40^{\circ}}+\frac{3 \operatorname{cosec} 80^{\circ}}{\sec 10^{\circ}}–2 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$?
Question : Find the value of the given expression. $\frac{4}{3} \tan^2 45^{\circ}+3 \cos^2 30^{\circ}-2 \sec^2 30^{\circ}-\frac{3}{4} \cot^2 60^{\circ}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile