Question : Simplify the following. $\frac{\sin^3 \alpha+\cos^3 \alpha}{\sin \alpha+\cos \alpha}$
Option 1: $1+\sin \alpha \cos \alpha$
Option 2: $\tan \alpha$
Option 3: $1-\sin \alpha \cos \alpha$
Option 4: $\sec \alpha$
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: $1-\sin \alpha \cos \alpha$
Solution : $\frac{\sin^3 \alpha+\cos^3 \alpha}{\sin \alpha+\cos \alpha}$ = $\frac{(\sin \alpha+\cos \alpha)(\sin^2 \alpha+\cos^2 \alpha–\sin \alpha\cos \alpha)}{\sin \alpha+\cos \alpha}$ = $(1–\sin \alpha\cos \alpha)$ [$\because \sin^2 \alpha+\cos^2 \alpha=1]$ Hence, the correct answer is $(1–\sin \alpha\cos \alpha)$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : Simplify the following expression. $\frac{\sin \theta - 2 \sin ^3 \theta}{2 \cos ^3 \theta - \cos \theta}$
Question : What is $\sin \alpha - \sin\beta$?
Question : Simplify: $\frac{\cos A}{1+\tan A}-\frac{\sin A}{1+\cot A}$
Question : Simplify $\frac{1 + \sin t}{4 - 4 \sin t} - \frac{1 - \sin t}{4 + 4 \sin t}$.
Question : Which of the following is equal to $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile