If \(\dfrac{1}{6}\) sin θ, cos θ, tan θ are in G.P, then the solution set of θ is (Here n ∈ ℕ)
\(2n\pi \pm \dfrac{\pi}{3}\)
Three quantities are in geometric progression when the square of the middle term equals the product of the outer terms; this gives a cubic in cos θ.
Step 1 — \(\cos^2\theta = \dfrac{1}{6}\sin\theta\tan\theta = \dfrac{\sin^2\theta}{6\cos\theta}\), so \(6\cos^3\theta = 1 - \cos^2\theta\).
Step 2 — With c = cos θ: \(6c^3 + c^2 - 1 = 0\), which factorises as \((2c-1)(3c^2+2c+1) = 0\).
Step 3 — The quadratic factor has discriminant \(4 - 12 \lt 0\), so the only real value is \(\cos\theta = \dfrac{1}{2}\).
Step 4 — The general solution of \(\cos\theta = \cos\dfrac{\pi}{3}\) is \(\theta = 2n\pi \pm \dfrac{\pi}{3}\); cos θ ≠ 0 there, so tan θ is defined.
The family with π/6 gives cos θ = √3/2, which does not satisfy the cubic. The form nπ + (−1)nπ/3 is the solution set of sin θ = √3/2 and includes 2π/3, where cos θ is negative. The family nπ + π/3 includes 4π/3, where cos θ = −1/2.
\(2n\pi \pm \dfrac{\pi}{3}\) option is correct.
Similar Questions
- If $\sum_{r=1}^{\infty} \tan^{-1}\left(\frac{1}{2r^2}\right) = a$, then $\tan a$ is equal to
- $\theta$ elimination from the equations $x^2 + y^2 = \frac{x \cos 3\theta + y \sin 3\theta}{\cos^3 \theta} = \frac{y \cos 3\theta - x \sin 3\theta}{\sin^3 \theta}$ will be
- The true set of values of '$K$' for which $\sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) = \frac{K\pi}{6}$ may have a solution is
- The general solution of the equation $\sin^{100} x - \cos^{100} x = 1$ is
- If for two real numbers $a, b$ with $|a| \le 1$ and $|b| \le 1$, $\frac{1}{3} + \frac{\sin^{-1} a + \sin^{-1} b}{4} + \frac{\left(\sin^{-1} a + \sin^{-1} b\right)^2}{16} + \frac{\left(\sin^{-1} a + \sin^{-1} b\right)^3}{64} + \dots = \frac{2(8-3\pi)}{3(16+3\pi)}$, then the value of $\sin^{-1}\left(a\sqrt{1-b^2} + b\sqrt{1-a^2}\right)$ is
- Let $g(x) = ax + b$, where $a < 0$ and $g$ is defined from $[1, 3]$ onto $[0, 2]$. Then the value of $\cot \left( \cos^{-1}(|\sin x| + |\cos x|) + \sin^{-1}(-|\cos x| - |\sin x|) \right)$ is equal to
The locus of points (x, y) in the plane satisfying sin2 x + sin2 y = 1 consists of
ABC is an isosceles triangle with an inscribed circle with centre O. Let P be the midpoint of BC. If AB = AC = 15 and BC = 10, then OP equals
If α1, α2, …, αn are in A.P. with common difference θ, then the sum of the series
sec α1 sec α2 + sec α2 sec α3 + …. + sec αn − 1 sec αn = k(tan αn − tan α1)
where k =
If A and B are acute angles such that sin A = sin2 B and 2 cos2 A = 3 cos2 B, then (A, B) =
Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
Let $\vec{a_k} = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot \theta_k) \hat{j}$, where $\theta_k = \frac{2^{k - 1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a_k}|^2}{\sum_{k=1}^n |\vec{b_k}|^2}$ is _____.
- If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
- Let the maximum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m}{n} \pi^2$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to __________.
