Question : The value of $\operatorname{cosec}^{2}60°+\sec^{2}60°– \cot^{2}60°+\tan^{2}30°$ will be:
Option 1: $5$
Option 2: $5\frac{1}{2}$
Option 3: $5\frac{1}{3}$
Option 4: $5\frac{2}{3}$
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: $5\frac{1}{3}$
Solution : Given: $\operatorname{cosec}^{2}60°+\sec^{2}60°– \cot^{2}60°+\tan^{2}30°$ We know that $\operatorname{cosec60°} = \frac{2}{\sqrt{3}}$ $\sec 60° = 2$ $\cot 60° = \frac{1}{\sqrt{3}}$ $\tan 30° = \frac{1}{\sqrt{3}}$ Substituting these values in the given expression, we get: = $(\frac{2}{\sqrt{3}})^{2} + 2^{2} – (\frac{1}{\sqrt{3}})^{2} +(\frac{1}{\sqrt{3}})^{2}$ = $\frac{4}{3}+4=\frac{16}{3}= 5\frac{1}{3}$ Hence, the correct answer is $5\frac{1}{3}$.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : The value of $\sqrt{\frac{1+\cos A}{1-\cos A}}$ is:
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
Question : If $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{1}{7}, \theta$ lies in first quadrant, then the value of $\frac{\operatorname{cosec} \theta+\cot ^2 \theta}{\operatorname{cosec} \theta-\cot ^2 \theta}$ is:
Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Question : If $\tan\alpha=2$, then the value of $\frac{\operatorname{cosec}^{2}\alpha-\sec^{2}\alpha}{\operatorname{cosec}^{2}\alpha+\sec^{2}\alpha}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile