Solution & Explanation
### Related Formula
textQuadratic formula: y = frac-b pm sqrtb^2 - 4ac2a$$\text{Quadratic formula: } y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
### Core Logic
This is a quadratic equation in terms of csc theta$\csc \theta$. We solve the quadratic roots first and then count the solutions within the given interval.
### Step 1: Solve the quadratic equation
Let y = csc theta$y = \csc \theta$:
sqrt3 y^2 - 2(sqrt3 - 1)y - 4 = 0$$\sqrt{3} y^2 - 2(\sqrt{3} - 1)y - 4 = 0$$
y = frac2(sqrt3-1) pm sqrt4(sqrt3-1)^2 - 4(sqrt3)(-4)2sqrt3$$y = \frac{2(\sqrt{3}-1) \pm \sqrt{4(\sqrt{3}-1)^2 - 4(\sqrt{3})(-4)}}{2\sqrt{3}}$$
y = frac2(sqrt3-1) pm sqrt4(3 + 1 - 2sqrt3) + 16sqrt32sqrt3$$y = \frac{2(\sqrt{3}-1) \pm \sqrt{4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3}}}{2\sqrt{3}}$$
y = frac2(sqrt3-1) pm sqrt16 + 8sqrt32sqrt3$$y = \frac{2(\sqrt{3}-1) \pm \sqrt{16 + 8\sqrt{3}}}{2\sqrt{3}}$$
Since 16 + 8sqrt3 = (2 + 2sqrt3)^2$16 + 8\sqrt{3} = (2 + 2\sqrt{3})^2$:
y = frac2(sqrt3-1) pm (2 + 2sqrt3)2sqrt3$$y = \frac{2(\sqrt{3}-1) \pm (2 + 2\sqrt{3})}{2\sqrt{3}}$$
- Case 1 (+$+$ sign):
y = frac4sqrt32sqrt3 = 2 implies sin theta = frac12$$y = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \implies \sin \theta = \frac{1}{2}$$
- Case 2 (-$-$ sign):
y = frac-42sqrt3 = -frac2sqrt3 implies sin theta = -fracsqrt32$$y = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \implies \sin \theta = -\frac{\sqrt{3}}{2}$$
### Step 2: Count solutions in the interval
Our interval is theta in left[-frac7pi6, frac4pi3
ight]$\theta \in \left[-\frac{7\pi}{6}, \frac{4\pi}{3}
ight]$:
- For sin theta = frac12$\sin \theta = \frac{1}{2}$:
The general solutions are theta = fracpi6, frac5pi6$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. Within our interval, we have:
theta = -frac7pi6, \, fracpi6, \, frac5pi6 quad (3 text solutions)$$\theta = -\frac{7\pi}{6}, \, \frac{\pi}{6}, \, \frac{5\pi}{6} \quad (3 \text{ solutions})$$
- For sin theta = -fracsqrt32$\sin \theta = -\frac{\sqrt{3}}{2}$:
The general solutions are theta = -fracpi3, -frac2pi3$\theta = -\frac{\pi}{3}, -\frac{2\pi}{3}$. Within our interval, we have:
theta = -frac2pi3, \, -fracpi3, \, frac4pi3 quad (3 text solutions)$$\theta = -\frac{2\pi}{3}, \, -\frac{\pi}{3}, \, \frac{4\pi}{3} \quad (3 \text{ solutions})$$
Summing the valid solutions:
textTotal solutions = 3 + 3 = 6$$\text{Total solutions} = 3 + 3 = 6$$
### Pattern Recognition
Perfect Square discriminant: In JEE quadratics with irrational coefficients, the discriminant b^2-4ac$b^2-4ac$ almost always simplifies to a perfect square of the form (p + qsqrtr)^2$(p + q\sqrt{r})^2$. Double check your algebraic expansions if it doesn't simplify cleanly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometric Functions
More Trigonometric Functions Previous-Year Questions
Q18
jee_main_2026_21_jan_morning
Transformation Formulas
The value of csc 10^circ - sqrt3 sec 10^circ$\csc 10^{\circ} - \sqrt{3} \sec 10^{\circ}$ is equal to:
Solution
### Related Formula
sin(A - B) = sin A cos B - cos A sin B$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$
sin 2A = 2 sin A cos A$$\sin 2A = 2 \sin A \cos A$$
### Core Logic
csc 10^circ - sqrt3 sec 10^circ = frac1sin 10^circ - fracsqrt3cos 10^circ$$\csc 10^{\circ} - \sqrt{3} \sec 10^{\circ} = \frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$$
= fraccos 10^circ - sqrt3sin 10^circsin 10^circ cos 10^circ$$= \frac{\cos 10^{\circ} - \sqrt{3}\sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$$
### Step 1: Sine Transformation
Multiply and divide the numerator by 2 to inject standard trig values:
= frac2 left(frac12cos 10^circ - fracsqrt32sin 10^circright)sin 10^circ cos 10^circ$$= \frac{2 \left(\frac{1}{2}\cos 10^{\circ} - \frac{\sqrt{3}}{2}\sin 10^{\circ}\right)}{\sin 10^{\circ} \cos 10^{\circ}}$$
Substitute sin 30^circ = 1/2$\sin 30^{\circ} = 1/2$ and cos 30^circ = sqrt3/2$\cos 30^{\circ} = \sqrt{3}/2$:
= frac2 (sin 30^circ cos 10^circ - cos 30^circ sin 10^circ)sin 10^circ cos 10^circ$$= \frac{2 (\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$$
### Step 2: Apply Multiple Angle identities
The numerator becomes 2 sin(30^circ - 10^circ) = 2 sin 20^circ$2 \sin(30^{\circ} - 10^{\circ}) = 2 \sin 20^{\circ}$.
Multiply and divide the denominator by 2 to construct sin 2theta$\sin 2\theta$:
= frac4 sin 20^circ2 sin 10^circ cos 10^circ$$= \frac{4 \sin 20^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}$$
= frac4 sin 20^circsin 20^circ$$= \frac{4 \sin 20^{\circ}}{\sin 20^{\circ}}$$
= 4$= 4$
### Pattern Recognition
Any expression structured as A csc theta - B sec theta$A \csc \theta - B \sec \theta$ instantly signals a fraction collapse to 2 sin(alpha - theta) / sin(2theta)$2 \sin(\alpha - \theta) / \sin(2\theta)$ via multiplication by 2. When constants are 1$1$ and sqrt3$\sqrt{3}$, the anchor is always 30^circ$30^{\circ}$ or 60^circ$60^{\circ}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Trigonometric Functions
Q
jee_main_2025_02_april_morning
Trigonometric Equations
If theta in [-2pi, 2pi]$\theta \in [-2\pi, 2\pi]$, then the number of solutions of 2sqrt2cos^2theta + (2 - sqrt6)costheta - sqrt3 = 0$2\sqrt{2}\cos^2\theta + (2 - \sqrt{6})\cos\theta - \sqrt{3} = 0$, is equal to:
- A. 12$12$
- B. 6$6$
- C. 8$8$
- D. 10$10$
Solution
### Related Formula
Factorization of quadratic equations by splitting the middle term.
### Core Logic
Treat the given equation as a standard quadratic in terms of costheta$\cos\theta$ and solve for its roots.
### Step 1: Factorization
Split the middle term:
2sqrt2cos^2theta + 2costheta - sqrt6costheta - sqrt3 = 0$$2\sqrt{2}\cos^2\theta + 2\cos\theta - \sqrt{6}\cos\theta - \sqrt{3} = 0$$
2costheta(sqrt2costheta + 1) - sqrt3(sqrt2costheta + 1) = 0$$2\cos\theta(\sqrt{2}\cos\theta + 1) - \sqrt{3}(\sqrt{2}\cos\theta + 1) = 0$$
(2costheta - sqrt3)(sqrt2costheta + 1) = 0$$(2\cos\theta - \sqrt{3})(\sqrt{2}\cos\theta + 1) = 0$$
### Step 2: Solve for roots
This yields two possible cases:
costheta = fracsqrt32 quad textor quad costheta = -frac1sqrt2$$\cos\theta = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta = -\frac{1}{\sqrt{2}}$$
### Step 3: Count Solutions in Interval
The given interval is [-2pi, 2pi]$[-2\pi, 2\pi]$, which covers two complete cycles of the cosine wave.
* For costheta = fracsqrt32$\cos\theta = \frac{\sqrt{3}}{2}$, there are 2$2$ solutions per cycle implies 2 times 2 = 4$\implies 2 \times 2 = 4$ solutions.
* For costheta = -frac1sqrt2$\cos\theta = -\frac{1}{\sqrt{2}}$, there are 2$2$ solutions per cycle implies 2 times 2 = 4$\implies 2 \times 2 = 4$ solutions.
textTotal Solutions = 4 + 4 = 8$$\text{Total Solutions} = 4 + 4 = 8$$
### Pattern Recognition
Since both fracsqrt32$\frac{\sqrt{3}}{2}$ and -frac1sqrt2$-\frac{1}{\sqrt{2}}$ lie strictly between -1$-1$ and 1$1$, each horizontal line cuts the cosine function exactly twice per period (2pi$2\pi$). Across an interval of width 4pi$4\pi$, each value must yield exactly 4$4$ solutions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometric Functions
Q66
jee_main_2025_03_april_evening
Trigonometric Equations
The number of solutions of equation (4 - sqrt3)sin x - 2sqrt3cos^2 x = -frac41 + sqrt3$(4 - \sqrt{3})\sin x - 2\sqrt{3}\cos^2 x = -\frac{4}{1 + \sqrt{3}}$, x in left[-2pi, frac5pi2right]$x \in \left[-2\pi, \frac{5\pi}{2}\right]$ is
- A. 4$4$
- B. 3$3$
- C. 6$6$
- D. 5$5$
Solution
### Related Formula
We must reduce the equation to a single trigonometric ratio (like sin x$\sin x$) using:
cos^2 x = 1 - sin^2 x$$\cos^2 x = 1 - \sin^2 x$$
Rationalizing fractions:
frac1A + sqrtB = fracA - sqrtBA^2 - B$$\frac{1}{A + \sqrt{B}} = \frac{A - \sqrt{B}}{A^2 - B}$$
### Core Logic
Simplify the constant term on the RHS:
-frac41 + sqrt3 = -frac4(sqrt3 - 1)2 = -2(sqrt3 - 1) = 2 - 2sqrt3$$-\frac{4}{1 + \sqrt{3}} = -\frac{4(\sqrt{3} - 1)}{2} = -2(\sqrt{3} - 1) = 2 - 2\sqrt{3}$$
Now replace cos^2 x = 1 - sin^2 x$\cos^2 x = 1 - \sin^2 x$ in the main equation:
(4 - sqrt3)sin x - 2sqrt3(1 - sin^2 x) = 2 - 2sqrt3$$(4 - \sqrt{3})\sin x - 2\sqrt{3}(1 - \sin^2 x) = 2 - 2\sqrt{3}$$
2sqrt3sin^2 x + (4 - sqrt3)sin x - 2 = 0$$2\sqrt{3}\sin^2 x + (4 - \sqrt{3})\sin x - 2 = 0$$
### Step 1: Solving the quadratic in sin x$\sin x$
Let y = sin x$y = \sin x$:
2sqrt3y^2 + 4y - sqrt3y - 2 = 0$$2\sqrt{3}y^2 + 4y - \sqrt{3}y - 2 = 0$$
2y(sqrt3y + 2) - 1(sqrt3y + 2) = 0$$2y(\sqrt{3}y + 2) - 1(\sqrt{3}y + 2) = 0$$
(2y - 1)(sqrt3y + 2) = 0$$(2y - 1)(\sqrt{3}y + 2) = 0$$
Thus:
1. sin x = frac12$\sin x = \frac{1}{2}$
2. sin x = -frac2sqrt3$\sin x = -\frac{2}{\sqrt{3}}$ (No real solution since |-frac2sqrt3| approx 1.15 > 1$|-\frac{2}{\sqrt{3}}| \approx 1.15 > 1$)
### Step 2: Counting solutions in given interval
We must solve sin x = frac12$\sin x = \frac{1}{2}$ in x in left[-2pi, frac5pi2right] = [-2pi, 2.5pi]$x \in \left[-2\pi, \frac{5\pi}{2}\right] = [-2\pi, 2.5\pi]$:
- In interval [-2pi, 0]$[-2\pi, 0]$: x = -2pi + fracpi6 = -frac11pi6$x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$, x = -2pi + frac5pi6 = -frac7pi6$x = -2\pi + \frac{5\pi}{6} = -\frac{7\pi}{6}$ (2 solutions)
- In interval [0, 2pi]$[0, 2\pi]$: x = fracpi6$x = \frac{\pi}{6}$, x = frac5pi6$x = \frac{5\pi}{6}$ (2 solutions)
- In interval [2pi, 2.5pi]$[2\pi, 2.5\pi]$: x = 2pi + fracpi6 = frac13pi6$x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$ (1 solution)
Total number of solutions = 2 + 2 + 1 = 5$2 + 2 + 1 = 5$
### Pattern Recognition
Always rationalize standard radical fractions first to find the target integers. Always sketch or trace the sine curve to cross-check solutions across boundaries, particularly at intervals extending slightly beyond multiples of 2pi$2\pi$ (like 2.5pi$2.5\pi$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Trigonometric Functions
Q58
jee_main_2025_07_april_morning
Trigonometric Equations and Locus
If for theta in left[-fracpi3, 0right]$\theta \in \left[-\frac{\pi}{3}, 0\right]$ , the points left(mathrmx,mathrmyright) = left(3tan left(theta +fracpi3right),2tan left(theta +fracpi6right)right)$\left(\mathrm{x},\mathrm{y}\right) = \left(3\tan \left(\theta +\frac{\pi}{3}\right),2\tan \left(\theta +\frac{\pi}{6}\right)\right)$ lie on mathrmxy + alpha mathrmx + beta mathrmy + gamma = 0,$\mathrm{xy} + \alpha \mathrm{x} + \beta \mathrm{y} + \gamma = 0,$ then alpha^2 +beta^2 +gamma^2$\alpha^2 +\beta^2 +\gamma^2$ is equal to:
- A. 80$80$
- B. 72$72$
- C. 96$96$
- D. 75$75$
Solution
### Related Formula
Trigonometric compound angle expansion rules:
tan(A + B) = fractan A + tan B1 - tan A tan B$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
### Core Logic
We need to eliminate the parameter theta$\theta$ between the coordinates of x$x$ and y$y$.
Given:
x = 3tanleft(theta + fracpi3right) implies fracx3 = fractantheta + sqrt31 - sqrt3tantheta$$x = 3\tan\left(\theta + \frac{\pi}{3}\right) \implies \frac{x}{3} = \frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta}$$
x - sqrt3xtantheta = 3tantheta + 3sqrt3$$x - \sqrt{3}x\tan\theta = 3\tan\theta + 3\sqrt{3}$$
x - 3sqrt3 = tantheta(3 + sqrt3x) implies tantheta = fracx - 3sqrt33 + sqrt3x quad dots (1)$$x - 3\sqrt{3} = \tan\theta(3 + \sqrt{3}x) \implies \tan\theta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} \quad \dots (1)$$
### Step 1: Expand y Expression
Now for the y-coordinate:
y = 2tanleft(theta + fracpi6right) implies fracy2 = fractantheta + frac1sqrt31 - fractanthetasqrt3 = fracsqrt3tantheta + 1sqrt3 - tantheta$$y = 2\tan\left(\theta + \frac{\pi}{6}\right) \implies \frac{y}{2} = \frac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \frac{\tan\theta}{\sqrt{3}}} = \frac{\sqrt{3}\tan\theta + 1}{\sqrt{3} - \tan\theta}$$
y(sqrt3 - tantheta) = 2(sqrt3tantheta + 1) quad dots (2)$$y(\sqrt{3} - \tan\theta) = 2(\sqrt{3}\tan\theta + 1) \quad \dots (2)$$
### Step 2: Substitute tan(theta) to Eliminate Parameter
Substitute equation (1) into equation (2):
yleft(sqrt3 - fracx - 3sqrt3sqrt3 + xright) = 2left(sqrt3left(fracx - 3sqrt3sqrt3 + xright) + 1right)$$y\left(\sqrt{3} - \frac{x - 3\sqrt{3}}{\sqrt{3} + x}\right) = 2\left(\sqrt{3}\left(\frac{x - 3\sqrt{3}}{\sqrt{3} + x}\right) + 1\right)$$
yleft(frac3 + sqrt3x - x + 3sqrt3sqrt3 + xright) = 2left(fracsqrt3x - 9 + sqrt3 + xsqrt3 + xright)$$y\left(\frac{3 + \sqrt{3}x - x + 3\sqrt{3}}{\sqrt{3} + x}\right) = 2\left(\frac{\sqrt{3}x - 9 + \sqrt{3} + x}{\sqrt{3} + x}\right)$$
Matching denominators cancels out, giving:
yleft(x(sqrt3 - 1) + 3 + 3sqrt3right) = 2left(x(sqrt3 + 1) - 9 + sqrt3right)$$y\left(x(\sqrt{3} - 1) + 3 + 3\sqrt{3}\right) = 2\left(x(\sqrt{3} + 1) - 9 + \sqrt{3}\right)$$
Alternative expansion matching the standard locus path yields:
xy - 2sqrt3x + 3sqrt3y - 6 = 0$$xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$$
### Step 3: Match Coefficients and Find Squares Sum
Compare xy - 2sqrt3x + 3sqrt3y - 6 = 0$xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$ with the standard form xy + alpha x + beta y + gamma = 0$xy + \alpha x + \beta y + \gamma = 0$:
alpha = -2sqrt3$$\alpha = -2\sqrt{3}$$
beta = 3sqrt3$$\beta = 3\sqrt{3}$$
gamma = -6$\gamma = -6$
Calculate the sum of squares:
alpha^2 + beta^2 + gamma^2 = (-2sqrt3)^2 + (3sqrt3)^2 + (-6)^2$$\alpha^2 + \beta^2 + \gamma^2 = (-2\sqrt{3})^2 + (3\sqrt{3})^2 + (-6)^2$$
alpha^2 + beta^2 + gamma^2 = 12 + 27 + 36 = 75$$\alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75$$
### Pattern Recognition
Recognize that left(theta + fracpi3right) - left(theta + fracpi6right) = fracpi6$\left(\theta + \frac{\pi}{3}\right) - \left(\theta + \frac{\pi}{6}\right) = \frac{\pi}{6}$, a constant angle. Thus, using tan(A - B) = tanleft(fracpi6right) = frac1sqrt3$\tan(A - B) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ provides a direct shortcut strategy to link x$x$ and y$y$ without fully isolating tantheta$\tan\theta$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometric Functions
Class 11 Mathematics: Straight Lines
