Question : What is the value of $\frac{\cos 37^{\circ}}{\sin 53^{\circ}}?$
Option 1: $ \frac{1}{2} $
Option 2: $\frac{1}{\sqrt{2}}$
Option 3: $0$
Option 4: $1$
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: $1$
Solution : $\frac{\cos 37^{\circ}}{\sin 53^{\circ}}$ $=\frac{\cos 37^{\circ}}{\sin (90^{\circ}-37^{\circ})}$ $=\frac{\cos 37^{\circ}}{\cos 37^{\circ}}$ $=1$ Hence, the correct answer is $1$.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : What will be the value of $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$?
Question : Find the value of $\frac{\cos^2 15^{\circ}-\sin^2 15^{\circ}}{\cos^2 145^{\circ}+\sin^2 145^{\circ}}$.
Question : The value of $\frac{4 \tan ^2 30^{\circ}+\sin ^2 30^{\circ} \cos ^2 45^{\circ}+\sec ^2 48^{\circ}-\cot ^2 42^{\circ}}{\cos 37^{\circ} \sin 53^{\circ}+\sin 37^{\circ} \cos 53^{\circ}+\tan 18^{\circ} \tan 72^{\circ}}$ is:
Question : Using $\cos (A+B)=\cos A \cos B-\sin A \sin B$, find the value of $\cos 75^\circ$.
Question : What is the value of $5 \sqrt{3} \cos 60^{\circ} \tan 30^{\circ}-3 \cos 0^{\circ}+3 \cos ^2 45^{\circ}+2 \sin ^2 60^{\circ}$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile