Question : The value of $\sqrt{\frac{1+\cos A}{1-\cos A}}$ is:
Option 1: $\sec A – \tan A$
Option 2: $\operatorname{cosec} A + \cot A$
Option 3: $\sec A + \tan A$
Option 4: $\operatorname{cosec} A – \cot A$
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: $\operatorname{cosec} A + \cot A$
Solution : Given, $\sqrt{\frac{1+\cos A}{1-\cos A}}$ $=\sqrt{\frac{1+\cos A}{1-\cos A}\times \frac{1+\cos A}{1+\cos A}}$ $=\sqrt{\frac{(1+\cos A)^2}{(1-\cos A)(1+\cos A)}}$ $=\sqrt{\frac{(1+\cos A)^2}{1-\cos^2A}}$ [Using $\small(a-b)(a+b)=a^2-b^2$] $=\sqrt{\frac{(1+\cos A)^2}{\sin^2A}}$ $=\frac{1+\cos A}{\sin A}$ $=\frac{1}{\sin A}+\frac{\cos A}{\sin A}$ $=\operatorname{cosec}A+\cot A$ Hence, the correct answer is $\operatorname{cosec}A+\cot A$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
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 : For any real values of $\theta, \sqrt{\frac{\sec\theta \:-\: 1}{\sec\theta \:+\: 1}}=$?
Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile