Trigonometric Ratios And Identities – JEE Main MathematicsPractice Questions & PYQs

Expressing $\csc 10^{\circ} - \sqrt{3} \sec 10^{\circ}$ as $\frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$ allows the terms to combine into the fraction $\frac{\cos 10^{\circ} - \sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$. Factoring 2 out of the numerator results in $2(\frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ})$, which simplifies via the sine subtraction formula to $2 \sin(30^{\circ} - 10^{\circ})$, or $2 \sin 20^{\circ}$. Given that the denominator is equivalent to $\frac{1}{2} \sin 20^{\circ}$ by the sine double-angle identity, the entire expression becomes $\frac{2 \sin 20^{\circ}}{\frac{1}{2} \sin 20^{\circ}}$, which simplifies to 4.

The expression $\frac{\cos^2 48^{\circ} - \sin^2 12^{\circ}}{\sin^2 24^{\circ} - \sin^2 6^{\circ}}$ simplifies effectively by applying the trigonometric product identities $\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)$ and $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$. Using these, the numerator transforms into $\cos(48^{\circ}+12^{\circ})\cos(48^{\circ}-12^{\circ})$, which is $\cos 60^{\circ} \cos 36^{\circ}$, while the denominator becomes $\sin(24^{\circ}+6^{\circ})\sin(24^{\circ}-6^{\circ})$, or $\sin 30^{\circ} \sin 18^{\circ}$. Since $\cos 60^{\circ}$ and $\sin 30^{\circ}$ are both $\frac{1}{2}$, they cancel each other out, reducing the entire fraction to $\frac{\cos 36^{\circ}}{\sin 18^{\circ}}$.

Substituting the exact values $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ yields the ratio $\frac{\sqrt{5}+1}{\sqrt{5}-1}$. Multiplying both the numerator and the denominator by the conjugate $\sqrt{5}+1$ results in $\frac{(\sqrt{5}+1)^2}{5-1}$, which expands to $\frac{6+2\sqrt{5}}{4}$ and simplifies to $\frac{3+\sqrt{5}}{2}$. Comparing this to $\frac{\alpha + \beta \sqrt{5}}{2}$ identifies the values $\alpha = 3$ and $\beta = 1$, resulting in $\alpha + \beta = 4$.

The sum of the roots of the quadratic equation is given by $\tan A + \tan B = 2$ and their product is given by $\tan A \tan B = -5$. Using the tangent addition formula, we find that $\tan(A+B)$ is determined by the expression
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1-(-5)} = \frac{1}{3}.$
Since the cosine of an angle relates the adjacent side to the hypotenuse in a right triangle, a triangle with legs of 3 and 1 possesses a hypotenuse of $\sqrt{3^2 + 1^2} = \sqrt{10}$, which implies $\cos(A+B) = \frac{3}{\sqrt{10}}$.

Applying the trigonometric identity $\sin^2\left(\frac{A+B}{2}\right) = \frac{1-\cos(A+B)}{2}$, the target expression becomes $20 \left(\frac{1-\cos(A+B)}{2}\right)$, which simplifies to $10(1 - \cos(A+B))$. Substituting the calculated cosine value results in $10(1 - \frac{3}{\sqrt{10}})$, which evaluates to $10 - \frac{30}{\sqrt{10}}$. Rationalizing the second term yields $10 - 3\sqrt{10}$.

To determine the value of $\tan 2\alpha$, treat the angle $2\alpha$ as the sum of $(\alpha+\beta)$ and $(\alpha-\beta)$, which allows the use of the tangent addition identity: $\tan 2\alpha = \tan[(\alpha+\beta)+(\alpha-\beta)] = \frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}$. First, find the tangent values for these individual angles using the given trigonometric ratios. Since $\cos(\alpha+\beta) = -\frac{1}{10}$ and $\alpha+\beta$ falls in the second quadrant, the sine is $\sqrt{1 - (-\frac{1}{10})^2} = \frac{3\sqrt{11}}{10}$, yielding $\tan(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$. For the second angle, given $\sin(\alpha-\beta) = \frac{3}{8}$, the cosine is $\sqrt{1 - (\frac{3}{8})^2} = \frac{\sqrt{55}}{8}$, leading to $\tan(\alpha-\beta) = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}}$, which can be written as $\frac{3}{\sqrt{5}\sqrt{11}}$.

Substituting these components into the identity creates the expression $\frac{-3\sqrt{11} + \frac{3}{\sqrt{5}\sqrt{11}}}{1 - (-3\sqrt{11})\left(\frac{3}{\sqrt{5}\sqrt{11}}\right)}$. Simplifying the numerator requires finding a common denominator of $\sqrt{55}$, resulting in $\frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}} = \frac{-33\sqrt{5} + 3}{\sqrt{55}} = \frac{3(1-11\sqrt{5})}{\sqrt{55}}$. The denominator simplifies to $1 + \frac{9}{\sqrt{5}}$, which is $\frac{\sqrt{5}+9}{\sqrt{5}}$. Combining these, the expression becomes $\frac{3(1-11\sqrt{5})}{\sqrt{11}\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}+9}$, where the $\sqrt{5}$ factors cancel to leave $\frac{3(1-11\sqrt{5})}{\sqrt{11}(9+\sqrt{5})}$. Comparing this to the form $\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$, it is evident that $r=11$ and $s=9$, making the final sum $r+s = 20$.

The trigonometric expression simplifies considerably by expanding the product terms and grouping them according to the standard addition and subtraction identities. Distributing the terms, the expression becomes $\sin \frac{15\theta}{2} \cos 8\theta + \sin \frac{15\theta}{2} \sin 8\theta + \cos \frac{15\theta}{2} \cos 8\theta - \cos \frac{15\theta}{2} \sin 8\theta$. Rearranging this allows us to group terms into $(\cos \frac{15\theta}{2} \cos 8\theta + \sin \frac{15\theta}{2} \sin 8\theta) + (\sin \frac{15\theta}{2} \cos 8\theta - \cos \frac{15\theta}{2} \sin 8\theta)$. Applying the identities for $\cos(A-B)$ and $\sin(A-B)$ where $A = 8\theta$ and $B = \frac{15\theta}{2}$, the expression reduces directly to $\cos(8\theta - \frac{15\theta}{2}) + \sin(\frac{15\theta}{2} - 8\theta)$, which is equivalent to $\cos \frac{\theta}{2} - \sin \frac{\theta}{2}$.

To find the numerical value, we calculate $\cos \theta$ using the given condition $\cot \theta = -\frac{1}{2\sqrt{2}}$ in the second quadrant. Using $1 + \cot^2 \theta = \csc^2 \theta$, we find $\csc^2 \theta = 1 + \frac{1}{8} = \frac{9}{8}$, so $\sin^2 \theta = \frac{8}{9}$ and $\sin \theta = \frac{2\sqrt{2}}{3}$. Consequently, $\cos \theta = \cot \theta \sin \theta = (-\frac{1}{2\sqrt{2}}) (\frac{2\sqrt{2}}{3}) = -\frac{1}{3}$. Applying half-angle identities where $\frac{\theta}{2}$ lies in the first quadrant, we have $\cos \frac{\theta}{2} = \sqrt{\frac{1+\cos \theta}{2}} = \sqrt{\frac{1-1/3}{2}} = \sqrt{\frac{1}{3}}$ and $\sin \frac{\theta}{2} = \sqrt{\frac{1-\cos \theta}{2}} = \sqrt{\frac{1+1/3}{2}} = \sqrt{\frac{2}{3}}$. Substituting these values into our simplified expression yields $\sqrt{\frac{1}{3}} - \sqrt{\frac{2}{3}}$, resulting in $\frac{1-\sqrt{2}}{\sqrt{3}}$.

Rearranging the given equation $\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1$ allows us to isolate the term $\frac{\sin^2 C}{\sin^2 A}$ as $1 - \frac{\tan(A-B)}{\tan A}$, which simplifies to $\frac{\tan A - \tan(A-B)}{\tan A}$. Applying the trigonometric identity for the difference of tangents, $\tan A - \tan(A-B)$ is equal to $\frac{\sin B}{\cos A \cos(A-B)}$. Substituting this back into the expression and noting that $\tan A = \frac{\sin A}{\cos A}$, the equation reduces to $\sin^2 C = \frac{\sin A \sin B}{\cos(A-B)}$.

Expanding $\cos(A-B)$ and dividing both the numerator and the denominator by $\cos A \cos B$ transforms the right side into $\frac{\tan A \tan B}{1 + \tan A \tan B}$. By utilizing the identity $\sin^2 C = \frac{\tan^2 C}{1 + \tan^2 C}$ and equating it to our derived expression, we obtain:
$\tan^2 C = \tan A \tan B$
This relationship demonstrates that the square of the middle term $\tan C$ is equal to the product of the terms $\tan A$ and $\tan B$, which confirms that $\tan A, \tan C, \tan B$ form a geometric progression.

The numerator $\sqrt{3} \csc 20^{\circ} - \sec 20^{\circ}$ can be expressed as $\frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$, which combines to form $\frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$. By scaling the numerator by a factor of 2, the expression $2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})$ simplifies using the sine angle subtraction identity to $2 \sin(60^{\circ} - 20^{\circ})$, or $2 \sin 40^{\circ}$. Since the denominator of this numerator fraction is $\sin 20^{\circ} \cos 20^{\circ}$, which equals $\frac{1}{2} \sin 40^{\circ}$, the entire numerator reduces to $\frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$.

The denominator consists of the product $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$. Substituting $\cos 60^{\circ} = \frac{1}{2}$ leaves us with $\frac{1}{2} \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}$. Applying the product identity $\cos \theta \cos 2\theta \cos 4\theta = \frac{\sin 8\theta}{8 \sin \theta}$ where $\theta = 20^{\circ}$ shows that $\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}$ equals $\frac{\sin 160^{\circ}}{8 \sin 20^{\circ}}$. Because $\sin 160^{\circ} = \sin 20^{\circ}$, this product simplifies to $\frac{1}{8}$. Multiplying this by the $\frac{1}{2}$ already present results in a denominator of $\frac{1}{16}$. Dividing the simplified numerator of 4 by the denominator of $\frac{1}{16}$ yields 64.

The entire expression simplifies by utilizing the trigonometric identity $\frac{\sin \theta}{\cos 3\theta} = \frac{1}{2}(\tan 3\theta - \tan \theta)$. This transformation is effective because it allows the sum $A$ to be expressed as a telescoping series, where the inner terms cancel one another out.

Substituting this identity into each component of $A$ yields $A = \frac{1}{2}(\tan 9^{\circ} - \tan 3^{\circ}) + \frac{1}{2}(\tan 27^{\circ} - \tan 9^{\circ}) + \frac{1}{2}(\tan 81^{\circ} - \tan 27^{\circ})$. When these terms are combined, the intermediate values $\tan 9^{\circ}$ and $\tan 27^{\circ}$ subtract to zero, simplifying the expression to $A = \frac{1}{2}(\tan 81^{\circ} - \tan 3^{\circ})$. Given that $B = \tan 81^{\circ} - \tan 3^{\circ}$, it follows that $A = \frac{1}{2}B$, which means the ratio $\frac{B}{A}$ results in 2.

The expression simplifies significantly upon expanding the terms and applying compound angle identities. Distributing the components yields $\sin 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2} + \cos 7x \cos \frac{13x}{2} - \cos 7x \sin \frac{13x}{2}$. By regrouping these terms, the expression can be written as:

$\left(\sin 7x \cos \frac{13x}{2} - \cos 7x \sin \frac{13x}{2}\right) + \left(\cos 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2}\right)$

This matches the standard identities for $\sin(A-B)$ and $\cos(A-B)$, which reduces the entire structure to $\sin \frac{x}{2} + \cos \frac{x}{2}$.

Squaring this reduced expression yields $\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}$. Applying the identities $\sin^2 \theta + \cos^2 \theta = 1$ and $2 \sin \theta \cos \theta = \sin 2\theta$ simplifies this to $1 + \sin x$.

Given $\cot x = \frac{5}{12}$ and the location of $x$ in the third quadrant, determining $\sin x$ involves the identity $1 + \cot^2 x = \csc^2 x$. This leads to $1 + \frac{25}{144} = \frac{169}{144}$, resulting in $\csc x = -\frac{13}{12}$ and consequently $\sin x = -\frac{12}{13}$. Substituting this value into the squared sum provides $1 - \frac{12}{13} = \frac{1}{13}$. Because $x$ lies within the interval $(\pi, \frac{3\pi}{2})$, the angle $\frac{x}{2}$ falls in the second quadrant where the sum of its sine and cosine is positive, resulting in a final value of $\frac{1}{\sqrt{13}}$.

Evaluating the product of sines relies on the following trigonometric identity:
$\sin(\theta) \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin(3\theta)$
Setting $\theta = 10^\circ$ allows the expression for $K$ to simplify to $\frac{1}{4} \sin(30^\circ)$, which yields $K = \frac{1}{8}$ since $\sin(30^\circ) = \frac{1}{2}$. Substituting this value into the target expression produces $\sin \left( \frac{10 \cdot \frac{1}{8} \cdot \pi}{3} \right)$, which simplifies to $\sin \left( \frac{5\pi}{12} \right)$. Evaluating this sine value using the angle addition formula results in $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$.

The expression $2(\cos^{8}\theta-\sin^{8}\theta)\sec2\theta$ simplifies by factoring the term $\cos^{8}\theta-\sin^{8}\theta$ into $(\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta)$. Using the identities $\cos^4\theta - \sin^4\theta = \cos 2\theta$ and $\cos^4\theta + \sin^4\theta = 1 - \frac{1}{2}\sin^2 2\theta$, the original equation becomes:
$a^2 = 2 - \sin^2 2\theta$
The condition $\tan^2\theta \neq 1$ ensures that $\cos 2\theta \neq 0$, which prevents $\sin^2 2\theta$ from being 1. Consequently, $\sin^2 2\theta$ must lie within the interval $[0, 1)$, constraining $a^2$ to the range $(1, 2]$. Because no integer square $a^2$ exists in the interval $(1, 2]$, there are no valid values for $a$ in the set $S$, which means $n(S) = 0$.

The equation $\sin x + \sin^2 x = 1$ implies $\sin x = 1 - \sin^2 x$, which simplifies to $\sin x = \cos^2 x$. Dividing both sides by $\cos x$ shows that $\tan x = \cos x$, which allows the expression to be rewritten entirely in terms of $\cos x$. Substituting $\cos x$ for every $\tan x$ term transforms the given expression into:
$2\cos^{12} x + 6\cos^{10} x + 6\cos^8 x + 2\cos^6 x = 2(\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x)$
Recognizing that $\sin^2 x = \cos^4 x$ allows the terms inside the parentheses to be identified as the expansion of $(\sin^2 x + \cos^2 x)^3$, which is $(\sin^2 x)^3 + 3(\sin^2 x)^2(\cos^2 x) + 3(\sin^2 x)(\cos^2 x)^2 + (\cos^2 x)^3$. Since $\sin^2 x + \cos^2 x = 1$, the expression simplifies to $2(1)^3$, resulting in a final value of 2.

The relationship $\sin^2 \theta + \cos^2 \theta = 1$ provides the basis for simplifying the given equation. Substituting $\cos^2 \theta = 1 - \sin^2 \theta$ leads to:
$10 \sin^4 \theta + 15(1 - \sin^2 \theta)^2 = 6$
Expanding the squared term results in $10 \sin^4 \theta + 15(1 - 2 \sin^2 \theta + \sin^4 \theta) = 6$, which simplifies to $25 \sin^4 \theta - 30 \sin^2 \theta + 9 = 0$. Factoring the quadratic expression as $(5 \sin^2 \theta - 3)^2 = 0$ provides $\sin^2 \theta = \frac{3}{5}$ and $\cos^2 \theta = \frac{2}{5}$.

The components of the required expression are $\csc^6 \theta = (\frac{1}{\sin^2 \theta})^3 = (\frac{5}{3})^3 = \frac{125}{27}$, $\sec^6 \theta = (\frac{1}{\cos^2 \theta})^3 = (\frac{5}{2})^3 = \frac{125}{8}$, and $\sec^8 \theta = (\frac{1}{\cos^2 \theta})^4 = (\frac{5}{2})^4 = \frac{625}{16}$. Substituting these values into $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ results in the calculation:
$\frac{27\left(\frac{125}{27}\right) + 8\left(\frac{125}{8}\right)}{16\left(\frac{625}{16}\right)} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5}$

Multiplying the numerator and the denominator by $\sin(\frac{\pi}{6})$, which is the difference between the angles $(\frac{\pi}{4} + \frac{r\pi}{6})$ and $(\frac{\pi}{4} + (r-1)\frac{\pi}{6})$, transforms the expression into a form suitable for a telescoping sum. Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, the summation simplifies to $2 \sum_{r=1}^{13} \left( \cot\left( \frac{\pi}{4} + (r-1)\frac{\pi}{6} \right) - \cot\left( \frac{\pi}{4} + \frac{r\pi}{6} \right) \right)$.

Because this is a telescoping series, consecutive terms cancel out, leaving only the first and last boundary terms: $2 \left( \cot\left( \frac{\pi}{4} \right) - \cot\left( \frac{\pi}{4} + \frac{13\pi}{6} \right) \right)$. Substituting the known trigonometric values, $\cot\left( \frac{\pi}{4} \right) = 1$ and $\cot\left( \frac{\pi}{4} + \frac{13\pi}{6} \right) = \cot\left( \frac{29\pi}{12} \right) = 2 - \sqrt{3}$, the expression simplifies to $2(1 - (2 - \sqrt{3}))$, which results in $2\sqrt{3} - 2$. Comparing this to the form $a\sqrt{3} + b$, we find $a = 2$ and $b = -2$, and consequently, $a^2 + b^2 = 4 + 4 = 8$.

The core of this problem relies on simplifying the trigonometric expression using standard product-to-sum identities and the triple angle formula. Recognizing that $4 \cos x \cos \left(\frac{\pi}{3}-x \right) \cos \left(\frac{\pi}{3}+x \right)$ simplifies to $\cos 3x$, the original function $f(x) = 6+16\cos x \cdot \cos \left(\frac{\pi}{3}-x \right) \cdot \cos \left(\frac{\pi}{3}+x \right) \cdot \sin 3x \cdot \cos 6x$ can be rewritten as $f(x) = 6 + 4 \cos 3x \sin 3x \cos 6x$. By applying the double angle identity $2 \sin \theta \cos \theta = \sin 2\theta$ in sequence, the expression $4 \sin 3x \cos 3x$ becomes $2 \sin 6x$, which then combines with $\cos 6x$ to yield $f(x) = 6 + \sin 12x$.

Since the sine function oscillates between $-1$ and $1$, the range of $f(x)$ is defined by $[6-1, 6+1]$, which is $[5, 7]$. Identifying the bounds as $\alpha = 5$ and $\beta = 7$, we seek the distance of the point $(5, 7)$ from the line $3x+4y+12=0$ using the perpendicular distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Substituting the known values, the calculation becomes $d = \frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}}$, which simplifies to $\frac{55}{5}$, resulting in a final distance of $11$ units.

The underlying relationship between $x$ and $y$ is determined by eliminating the parameter $\theta$ using trigonometric addition formulas. By setting $\varphi = \theta + \frac{\pi}{6}$, we can express the coordinates as $y = 2 \tan \varphi$ and $x = 3 \tan\left(\varphi + \frac{\pi}{6}\right)$. Using the addition formula for tangent, the expression for $x$ becomes $x = 3 \left( \frac{\tan \varphi + \frac{1}{\sqrt{3}}}{1 - \tan \varphi \cdot \frac{1}{\sqrt{3}}} \right)$. Substituting $\tan \varphi = \frac{y}{2}$ into this equation results in $x = 3 \left( \frac{\frac{y}{2} + \frac{1}{\sqrt{3}}}{1 - \frac{y}{2\sqrt{3}}} \right)$, which simplifies to $x = 3 \left( \frac{y\sqrt{3} + 2}{2\sqrt{3} - y} \right)$. Clearing the denominator gives $x(2\sqrt{3} - y) = 3y\sqrt{3} + 6$, leading to $2\sqrt{3}x - xy = 3\sqrt{3}y + 6$. Rearranging this expression into the form $xy + \alpha x + \beta y + \gamma = 0$ yields $xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0$. Comparing this result with the given equation, we identify the coefficients as $\alpha = -2\sqrt{3}$, $\beta = 3\sqrt{3}$, and $\gamma = 6$. Finally, calculating the required sum of squares gives $\alpha^2 + \beta^2 + \gamma^2 = (-2\sqrt{3})^2 + (3\sqrt{3})^2 + 6^2 = 12 + 27 + 36 = 75$.

Distributing the $\sin 70^\circ$ term across the bracket simplifies the product $\sin 70^\circ \cot 70^\circ$ into $\cos 70^\circ$, which leaves the expression as $\cot 10^\circ \cos 70^\circ - \sin 70^\circ$. Converting the cotangent term into sine and cosine components yields $\frac{\cos 10^\circ \cos 70^\circ - \sin 70^\circ \sin 10^\circ}{\sin 10^\circ}$. The numerator now perfectly matches the cosine sum identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$, simplifying to $\cos(10^\circ + 70^\circ)$, or $\cos 80^\circ$. Since $\cos 80^\circ$ is equal to $\sin 10^\circ$, the entire fraction becomes $\frac{\sin 10^\circ}{\sin 10^\circ}$, resulting in 1.

Substituting $e^{\sin x} = t$ where $t > 0$ transforms the expression into the equation $t - \frac{2}{t} = 2$, which rearranges into the quadratic form $t^2 - 2t - 2 = 0$. Solving for $t$ yields $1 \pm \sqrt{3}$, and because $t$ must be positive, we retain only $t = 1 + \sqrt{3} \approx 2.73$. Since $e^{\sin x} = 1 + \sqrt{3}$, taking the natural logarithm on both sides results in $\sin x = \ln(1 + \sqrt{3})$. Given that $1 + \sqrt{3} > e$ (where $e \approx 2.718$), it follows that $\ln(1 + \sqrt{3}) > 1$. Because the range of the sine function is restricted to the interval $[-1, 1]$, the equation $\sin x > 1$ has no real roots, resulting in 0 solutions.

Starting with the expansion of the sine addition and subtraction formulas allows us to express the original equation as $3 \sin\alpha \cos\beta + 3 \cos\alpha \sin\beta = 2 \sin\alpha \cos\beta - 2 \cos\alpha \sin\beta$. Collecting the terms involving $\sin\alpha \cos\beta$ on the left side and those with $\cos\alpha \sin\beta$ on the right leads to $\sin\alpha \cos\beta = -5 \cos\alpha \sin\beta$. To isolate the tangent functions, dividing both sides by $\cos\alpha \cos\beta$ transforms the equation into $\frac{\sin\alpha}{\cos\alpha} = -5 \frac{\sin\beta}{\cos\beta}$, which is equivalent to $\tan\alpha = -5 \tan\beta$. Comparing this derived relationship to the definition $\tan\alpha = k \tan\beta$ identifies the value of $k$ as $-5$.