Question:medium
The value of $\frac{\sqrt{3}(\sin 40^{\circ}+\sin 20^{\circ})}{\cos 40^{\circ}+\cos 20^{\circ}}$ is
The value of $\frac{\sqrt{3}(\sin 40^{\circ}+\sin 20^{\circ})}{\cos 40^{\circ}+\cos 20^{\circ}}$ is
Show Hint
Logic Tip: Notice that $\frac{\sin A + \sin B}{\cos A + \cos B}$ will always simplify neatly to $\tan\left(\frac{A+B}{2}\right)$. Using this shortcut, the expression instantly becomes $\sqrt{3} \cdot \tan(30^{\circ}) = 1$.
Updated On: Apr 27, 2026
- 1
- $\sqrt{3}$
- $\sqrt{2}$
- $\frac{1}{2}$
- 0
Show Solution
The Correct Option is A
Solution and Explanation
Was this answer helpful?
0
Top Questions on Trigonometry
- If \( \tan^{-1}\left(\frac{1}{1+1\cdot2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}(x) \), then \( x \) is equal to:
- BITSAT - 2024
- Mathematics
- Trigonometry
- If \[ y = \tan^{-1} \left( \frac{1}{x^2 + x + 1} \right) + \tan^{-1} \left( \frac{1}{x^2 + 3x + 3} \right) + \tan^{-1} \left( \frac{1}{x^2 + 5x + 7} \right) + \cdots { (to n terms)} \], then \(\frac{dy}{dx}\) is:
- BITSAT - 2024
- Mathematics
- Trigonometry
- Let \(A\), \(B\) and \(C\) are the angles of a triangle and \(\tan \frac{A}{2} = 1/3\), \(\tan \frac{B}{2} = \frac{2}{3}\). Then, \(\tan \frac{C}{2}\) is equal to:
- BITSAT - 2024
- Mathematics
- Trigonometry
- The sum of all values of \(x\) in \([0, 2\pi]\), for which \(x + \sin(2x) + \sin(3x) + \sin(4x) = 0\) is equal to:
- BITSAT - 2024
- Mathematics
- Trigonometry
- Want to practice more? Try solving extra ecology questions todayView All Questions
