Question : If $\tan 40^{\circ}=\alpha$, then find $\frac{\tan 320^{\circ}-\tan 310^{\circ}}{1+\tan 320^{\circ} \cdot \tan 310^{\circ}}$.
Option 1: $\frac{1-\alpha^2}{\alpha}$
Option 2: $\frac{1+\alpha^2}{2 \alpha}$
Option 3: $\frac{1-\alpha^2}{2 \alpha}$
Option 4: $\frac{1+\alpha^2}{a}$
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: $\frac{1-\alpha^2}{2 \alpha}$
Solution : $\tan 40^{\circ}=\alpha$ $\frac{\tan 320^{\circ}-\tan 310^{\circ}}{1+\tan 320^{\circ} \cdot \tan 310^{\circ}}$ $= \frac{\tan(360°-40°) - \tan(270°+40°)}{1+\tan(360°-40°)\cdot\tan(270°+40°)}$ $= \frac{-\tan 40° + \cot 40°}{1-\tan40° \cdot\cot 40°}$ $= \frac{-\alpha + \frac{1}{\alpha}}{1+\alpha×\frac{1}{\alpha}}$ $= \frac{-\alpha^2 + 1}{2\alpha}$ $=\frac{1-\alpha^2}{2\alpha}$ Hence, the correct answer is $\frac{1-\alpha^2}{2\alpha}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x\cos^{2}30^{\circ}\cdot \sin60^{\circ}=\frac{\tan^{2}45^{\circ}\cdot \sec60^{\circ}}{\operatorname{cosec}60^{\circ}}$, then the value of $x$ is:
Question : If $\tan (\alpha+\beta)=a, \tan (\alpha-\beta)=b$, then the value of $\tan 2 \alpha$ is:
Question : $\cos ^2 35^{\circ}+\cos 55^{\circ} \cdot \sin 35^{\circ}+\frac{\tan 34^{\circ}}{\cot 56^{\circ}}=$________.
Question : If $x\sin ^{2}60^{\circ}-\frac{3}{2}\sec 60^{\circ}\tan^{2}30^{\circ}+\frac{4}{5}\sin ^{2}45^{\circ}\tan ^{2}60^{\circ}=0$, then $x$ is:
Question : What is the value of $\frac{\tan 45^{\circ}-\tan 15^{\circ}}{1+\tan 45^{\circ} \tan 15^{\circ}}$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile