If α, -π/2

Q: Wait, why do we square the equation? Doesn't that introduce fake solutions?

You are absolutely right to be cautious! Squaring an equation like $A = B$ to get $A^2 = B^2$ can indeed introduce extraneous roots because it essentially solves for both $A = B$ and $A = -B$. That is exactly why we must verify our final answers against the original equation or the domain constraints. It is a standard, powerful technique in trigonometry, provided you remember to perform that final check.

Q: Why do we divide by $\cos \theta$? What if $\cos \theta = 0$?

That is a brilliant question. Before dividing by $\cos \theta$, we should always check if $\cos \theta = 0$ could be a solution. If $\cos \theta = 0$, then $\sin \theta$ must be $\pm 1$. Substituting this into $4 \cos \theta + 5 \sin \theta = 1$ gives $5(\pm 1) = 1$, which is impossible. Therefore, $\cos \theta$ cannot be zero, and dividing by it is perfectly safe.

Q: How do I know which root to reject?

Great question. We use the domain constraint $\alpha \in (-\frac{\pi}{2}, \frac{\pi}{2})$. In this interval, $\cos \alpha$ is always positive. Since $\sec \alpha = \frac{1}{\cos \alpha}$, $\sec \alpha$ must also be positive. Looking at our transformed equation $4 + 5 \tan \alpha = \sec \alpha$, the left side must be positive. By testing both roots, you will find that one makes the expression negative, which violates our condition. That is the root you reject!

Question

If $\alpha, -\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ is the solution of $4 \cos 	heta + 5 \sin 	heta = 1$, then the value of $	an \alpha$ is

Options:

(A) $\frac{10 - \sqrt{10}}{6}$

(B) $\frac{10 - \sqrt{10}}{12}$

(C) $\frac{\sqrt{10} - 10}{12}$

(D) $\frac{\sqrt{10} - 10}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a unit circle, looking at the intersection of a line and a curve. The problem before us, $4 \cos 	heta + 5 \sin 	heta = 1$, is not just a collection of symbols; it is a geometric dance.

If we let $x = \cos 	heta$ and $y = \sin 	heta$, we are essentially finding where the line $4x + 5y = 1$ cuts through the unit circle $x^2 + y^2 = 1$. This is the heart of the problem.

We are looking for the angle $\alpha$ that satisfies this, restricted to the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

### The Algebraic Transformation

To solve this, we need to speak the language of $	an 	heta$. We achieve this by dividing the entire equation by $\cos 	heta$:

$$4 + 5 	an 	heta = \sec 	heta$$

This is a beautiful step because it brings us closer to a single variable. To eliminate $\sec 	heta$, we utilize the identity $\sec^2 	heta = 1 + 	an^2 	heta$ and square both sides:

$$(4 + 5 	an 	heta)^2 = \sec^2 	heta$$

This is the turning point. We have traded a simple equation for a quadratic one, gaining the power to solve for $	an 	heta$ directly.

### The Quadratic Masterclass

Now, let us expand the left side:

$$16 + 40 	an 	heta + 25 	an^2 	heta = 1 + 	an^2 	heta$$

Bringing everything to one side, we obtain the following quadratic equation:

$$24 	an^2 	heta + 40 	an 	heta + 15 = 0$$

Using the quadratic formula $	an 	heta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 24$, $b = 40$, and $c = 15$, we calculate the discriminant:

$$D = 40^2 - 4(24)(15) = 1600 - 1440 = 160$$

Thus, the roots are:

$$	an 	heta = \frac{-40 \pm \sqrt{160}}{48} = \frac{-40 \pm 4\sqrt{10}}{48} = \frac{-10 \pm \sqrt{10}}{12}$$

### The Final Filter: The Domain Constraint

We have two potential values for $	an \alpha$. We are given $\alpha \in (-\frac{\pi}{2}, \frac{\pi}{2})$, which implies $\cos \alpha > 0$. Consequently, $\sec \alpha = \frac{1}{\cos \alpha}$ must be positive.

Our equation $4 + 5 	an \alpha = \sec \alpha$ implies that $4 + 5 	an \alpha$ must be greater than zero. Let us test our roots:

For $	an \alpha = \frac{-10 - \sqrt{10}}{12}$, the expression $4 + 5(\frac{-10 - \sqrt{10}}{12})$ is clearly negative. We reject this root.

For $	an \alpha = \frac{-10 + \sqrt{10}}{12}$, the expression becomes:

$$4 + 5\left(\frac{-10 + \sqrt{10}}{12}ight) = \frac{48 - 50 + 5\sqrt{10}}{12} = \frac{5\sqrt{10} - 2}{12}$$

Since this value is positive, it is our valid solution. The final value is **$	an \alpha = \frac{\sqrt{10} - 10}{12}$**.

Animated Solution for Mathematics - Trigonometry: If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 cos θ + 5 sin θ = 1$ , then the value of $tan α$ is

Select Answer:

Visualized Solution

Visualizing the Problem

Converting to $tan θ$

Squaring Both Sides

Using the Identity $sec^{2} θ = 1 + tan^{2} θ$

Expanding the Left Side

Forming the Quadratic Equation

Applying the Quadratic Formula

Calculating the Discriminant

Simplifying the Roots

Checking the Domain Constraint

Testing the First Root

Testing the Second Root

Final Conclusion

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Algebraic Transformation

The Quadratic Masterclass

The Final Filter

Similar Questions

Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 cos θ - 3 sin θ = 1$ . Then $cos θ$ is equal to :

Find all values of $θ$ in the interval $(- π /2, π /2)$ satisfying the equation $(1 - tan θ) (1 + tan θ) sec^{2} θ + 2^{t a n^{2} θ} = 0$ .

Let $S = {θ \in [0, 2 π) : tan (π cos θ) + tan (π sin θ) = 0}$ . Then $\sum_{θ \in S} sin^{2} (θ + \frac{π}{4})$ is equal to

General value of $θ$ satisfying the equation $tan^{2} θ + sec 2 θ = 1$ is ..........

The sum of solutions of the equation $\frac{c o s x}{1 + s i n x} = ∣ tan 2 x ∣, x \in (- \frac{π}{2}, \frac{π}{2}) - {\frac{π}{4}, - \frac{π}{4}}$ is :

The number of values of $θ$ in the interval $(- π /2, π /2)$ such that $θ \neq = \frac{nπ}{5}$ for $n = 0, \pm 1, \pm 2$ and $tan θ = cot 5 θ$ as well as $sin 2 θ = cos 4 θ$ is ____.

If $θ_{1}$ and $θ_{2}$ be respectively the smallest and the largest values of $θ$ in $(0, 2 π) - {π}$ which satisfy the equation, $2 cot^{2} θ - \frac{5}{s i n θ} + 4 = 0$ , then $\int_{θ_{1}}^{θ_{2}} cos^{2} 3 θ d θ$ is equal to :

The sum of all values of $θ \in (0, \frac{π}{2})$ satisfying $sin^{2} 2 θ + cos^{4} 2 θ = \frac{3}{4}$ is :

Let $S = {θ \in [- π, π] - {\pm \frac{π}{2}} : sin θ tan θ + tan θ = sin 2 θ}$ . If $T = \sum_{θ \in S} cos 2 θ$ , then $T + n (S)$ is equal to

All possible values of $θ \in [0, 2 π]$ for which $sin 2 θ + tan 2 θ > 0$ lie in:

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α,−2π​<α<2π​ is the solution of 4cosθ+5sinθ=1, then the value of tanα is