Question : Find the value of $\tan (–1125°)$.
Option 1: $1$
Option 2: $\frac{1}{2}$
Option 3: $–1$
Option 4: $0$
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: $–1$
Solution : The angle $1125°$ is co-terminal to the $45°$ angle and also tangent is a periodic function. So, $\tan 1125°= \tan (6\times 180°+45°) = \tan 45°=1$ $\therefore \tan (–1125°) = – \tan 1125° = –1$ Hence, the correct answer is –1.
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 tan 25° tan 35° tan 45° tan 55° tan 65° is:
Question : Using trigonometric formulas, find the value of $(\frac{\sin (x-y)}{\sin (x+y)})(\frac{\tan x+\tan y}{\tan x-\tan y})$
Question : If $7\sin^2\ \theta+3\cos^2\ \theta=4, (0°<\theta<90°)$, then find the value of $\tan\theta$:
Question : The value of $\operatorname{cosec}^{2}60°+\sec^{2}60°– \cot^{2}60°+\tan^{2}30°$ will be:
Question : Using $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, find the value of $\tan 15°$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile