General value of θ satisfying the equation tan² θ + sec 2θ = 1 is ..........

Q: Wait, why do we reject the negative root here? Or do we?

That is a fantastic question! In this problem, we do not actually reject any roots based on sign. When we arrive at $\tan^2 \theta = 3$, we take the square root to get $\tan \theta = \pm \sqrt{3}$. Both positive and negative values are valid because the tangent function covers all real values. We simply account for both by using the $\pm$ sign in our general solution.

Q: Why can't we just use energy conservation... I mean, identity $\sec^2 2\theta = 1 + \tan^2 2\theta$?

I love the enthusiasm! However, using $\sec^2 2\theta = 1 + \tan^2 2\theta$ would introduce $\tan^2 2\theta$, which keeps the angle as $2\theta$. Our goal is to simplify the equation, and converting everything to $\tan \theta$ is the most direct path to a solvable polynomial.

Q: Is there a domain restriction I should worry about?

Yes, absolutely! Whenever you have a denominator, you must ensure it is not zero. Here, our denominator is $(1-\tan^2 \theta)$. This means $\tan^2 \theta \neq 1$, so $\tan \theta \neq \pm 1$. This implies $\theta \neq n\pi \pm \frac{\pi}{4}$. Our final solutions, $\theta = n\pi$ and $\theta = n\pi \pm \frac{\pi}{3}$, do not conflict with this, so they are all valid.

Question

General value of $	heta$ satisfying the equation $	an^2 	heta + \sec 2	heta = 1$ is ..........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a trigonometric equation; we are embarking on a journey of simplification and elegance.

When you look at the equation $	an^2 	heta + \sec 2	heta = 1$, it might seem like a chaotic mess of different functions and angles. But remember, in mathematics, chaos is often just order waiting to be discovered.

Our mission is to bring harmony to this equation by unifying everything into a single language: the language of $	an 	heta$.

### The Bridge of Identities

The primary obstacle here is the mismatch between $	heta$ and $2	heta$. We cannot easily combine terms when they are speaking different languages. We need a bridge.

That bridge is the double-angle identity for cosine:

$$ \cos 2	heta = \frac{1-	an^2 	heta}{1+	an^2 	heta} $$

This identity is a masterpiece of trigonometry, allowing us to leap from the double angle $2	heta$ back to the single angle $	heta$. Since we have $\sec 2	heta$ in our equation, we simply take the reciprocal of this identity:

$$ \sec 2	heta = \frac{1+	an^2 	heta}{1-	an^2 	heta} $$

### The Algebraic Dance

With our identity in hand, we substitute it back into the original equation:

$$ 	an^2 	heta + \frac{1+	an^2 	heta}{1-	an^2 	heta} = 1 $$

We can clear the path by multiplying the entire equation by the denominator, $(1 - 	an^2 	heta)$. This yields:

$$ 	an^2 	heta (1 - 	an^2 	heta) + (1 + 	an^2 	heta) = 1 - 	an^2 	heta $$

Expanding this, we get:

$$ 	an^2 	heta - 	an^4 	heta + 1 + 	an^2 	heta = 1 - 	an^2 	heta $$

Notice the magic happening here? The $1$ on both sides cancels out beautifully, leaving us with:

$$ 2	an^2 	heta - 	an^4 	heta = -	an^2 	heta $$

Rearranging this, we arrive at the elegant form:

$$ 3	an^2 	heta - 	an^4 	heta = 0 $$

### The Bifurcation of Solutions

We have arrived at the climax of our journey. Factoring out $	an^2 	heta$, we get:

$$ 	an^2 	heta (3 - 	an^2 	heta) = 0 $$

This splits our problem into two distinct, manageable cases.

**Case 1:** $	an^2 	heta = 0$, which implies $	an 	heta = 0$. On the unit circle, this occurs at every integer multiple of $\pi$, so $	heta = n\pi$.

**Case 2:** $3 - 	an^2 	heta = 0$, which means $	an^2 	heta = 3$. We know that $	an(\frac{\pi}{3}) = \sqrt{3}$, so $	an^2(\frac{\pi}{3}) = 3$.

The general solution for $	an^2 	heta = 	an^2 \alpha$ is $	heta = n\pi \pm \alpha$. Substituting $\alpha = \frac{\pi}{3}$, we get $	heta = n\pi \pm \frac{\pi}{3}$.

### Final Harmony

By combining these two cases, we reach our final destination. The general solution is:

**$	heta = n\pi$ and $	heta = n\pi \pm \frac{\pi}{3}$**

(where $n$ is any integer). We have successfully navigated the complexity, cleared the denominators, and found the hidden symmetry. Remember, every time you face a daunting equation, look for the identity that bridges the gap.

Animated Solution for Mathematics - Trigonometry: General value of $θ$ satisfying the equation $tan^{2} θ + sec 2 θ = 1$ is ..........

Visualized Solution

Introduction to the Equation

Applying the Double Angle Identity

Expressing $sec 2 θ$

Substituting into the Equation

Clearing the Denominator

Expanding and Simplifying

Factoring the Equation

Case 1: Solving for $tan θ = 0$

Case 2: Solving for $tan^{2} θ = 3$

Final Conclusion and Summary

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Bridge of Identities

The Algebraic Dance

The Bifurcation of Solutions

Final Harmony

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } General value of θ satisfying the equation tan2θ+sec2θ=1 is ..........