Question : If $\tan ^2 \alpha=3+Q^2$, then $\sec \alpha+\tan ^3 \alpha \operatorname{cosec} \alpha=?$
Option 1: $\left(3+Q^2\right)^{\frac{3}{2}}$
Option 2: $\left(7+Q^2\right)^{\frac{3}{2}}$
Option 3: $\left(5-Q^2\right)^{\frac{3}{2}}$
Option 4: $\left(4+Q^2\right)^{\frac{3}{2}}$
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: $\left(4+Q^2\right)^{\frac{3}{2}}$
Solution : Consider, $\sec \alpha+\tan ^3 \alpha \operatorname{cosec} \alpha$ = $\frac{1}{\cos\alpha}+\frac{\sin^3\alpha}{\cos^3\alpha}\times\frac{1}{\sin\alpha}$ = $\frac{1}{\cos\alpha}+\frac{\sin^2\alpha}{\cos^3\alpha}$ = $\frac{\cos^2\alpha+\sin^2\alpha}{\cos^3\alpha}$ = $\frac{1}{\cos^3\alpha}$ [Using $\cos^2\alpha+\sin^2\alpha = 1$] ....................(1) Now consider, $\tan ^2 \alpha=3+Q^2$ ⇒ $\tan\alpha = \sqrt{3+Q^2}$ Let base ($b$) = 1 and perpendicular ($p$) = $\sqrt{3+Q^2}$ Applying Pythagoras Theorem, $h^2=p^2+b^2$ [where $h,p,$ and $b$ are hypotenuse, perpendicular and base respectively.] ⇒ $h=\sqrt{1+3+Q^2}$ ⇒ $h=\sqrt{4+Q^2}$ ⇒ $\cos\alpha = \frac{b}{h}=\frac{1}{\sqrt{4+Q^2}}$ Putting in (1), we get, $\frac{1}{\cos^3\alpha}=\frac{1}{(\frac{1}{\sqrt{4+Q^2}})^3} = (4+Q^2)^{\frac{3}{2}}$ Hence, the correct answer is $(4+Q^2)^{\frac{3}{2}}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
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:
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 : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : If $\sec \theta+\tan \theta=5, (\theta \neq 0)$, then $\sec \theta$ is equal to:
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:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile