Understanding the Trigonometric Expression

The question asks us to simplify and find the value of the expression: $(\sec\theta + \tan\theta)^2 + \frac{1 + \operatorname{cosec}\theta}{1 - \operatorname{cosec}\theta}$, given that $0^\circ < \theta < 90^\circ$. This range for $\theta$ means that $\sin\theta$, $\cos\theta$, $\tan\theta$, $\sec\theta$, and $\operatorname{cosec}\theta$ are all positive and well-defined (since $\sin\theta \ne 1$ and $\cos\theta \ne 0$).

Step-by-Step Simplification

Let's simplify each part of the expression separately.

Simplifying the First Term: $(\sec\theta + \tan\theta)^2$

We know the basic trigonometric identities:

• $\sec\theta = \frac{1}{\cos\theta}$
• $\tan\theta = \frac{\sin\theta}{\cos\theta}$

Substitute these into the first term:

$(\sec\theta + \tan\theta)^2 = \left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right)^2$

Combine the terms inside the parenthesis:

$= \left(\frac{1 + \sin\theta}{\cos\theta}\right)^2$

Square the numerator and the denominator:

$= \frac{(1 + \sin\theta)^2}{\cos^2\theta}$

Using the identity $\cos^2\theta = 1 - \sin^2\theta$:

$= \frac{(1 + \sin\theta)^2}{1 - \sin^2\theta}$

The denominator $1 - \sin^2\theta$ is a difference of squares, which can be factored as $(1 - \sin\theta)(1 + \sin\theta)$.

$= \frac{(1 + \sin\theta)(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}$

Since $0^\circ < \theta < 90^\circ$, $\sin\theta \ne 1$, so we can cancel the $(1 + \sin\theta)$ term from numerator and denominator:

$= \frac{1 + \sin\theta}{1 - \sin\theta}$

So, the first term simplifies to $\frac{1 + \sin\theta}{1 - \sin\theta}$.

Simplifying the Second Term: $\frac{1 + \operatorname{cosec}\theta}{1 - \operatorname{cosec}\theta}$

We know the basic trigonometric identity:

• $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$

Substitute this into the second term:

$\frac{1 + \operatorname{cosec}\theta}{1 - \operatorname{cosec}\theta} = \frac{1 + \frac{1}{\sin\theta}}{1 - \frac{1}{\sin\theta}}$

Find a common denominator for the numerator and the denominator:

$= \frac{\frac{\sin\theta + 1}{\sin\theta}}{\frac{\sin\theta - 1}{\sin\theta}}$

Multiply the numerator by the reciprocal of the denominator:

$= \frac{\sin\theta + 1}{\sin\theta} \times \frac{\sin\theta}{\sin\theta - 1}$

Cancel the $\sin\theta$ terms (since $0^\circ < \theta < 90^\circ$, $\sin\theta \ne 0$):

$= \frac{\sin\theta + 1}{\sin\theta - 1}$

We can rewrite the denominator $\sin\theta - 1$ as $-(1 - \sin\theta)$.

$= \frac{1 + \sin\theta}{-(1 - \sin\theta)}$

$= - \frac{1 + \sin\theta}{1 - \sin\theta}$

So, the second term simplifies to $- \frac{1 + \sin\theta}{1 - \sin\theta}$.

Combining the Simplified Terms

Now, add the simplified first term and the simplified second term:

$(\sec\theta + \tan\theta)^2 + \frac{1 + \operatorname{cosec}\theta}{1 - \operatorname{cosec}\theta} = \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right) + \left(- \frac{1 + \sin\theta}{1 - \sin\theta}\right)$

$= \frac{1 + \sin\theta}{1 - \sin\theta} - \frac{1 + \sin\theta}{1 - \sin\theta}$

This is a term minus itself, which equals 0.

$= 0$

The value of the entire expression is 0.

Verification with Options

The calculated value is 0, which matches Option 1.

Conclusion

By simplifying both parts of the trigonometric expression using fundamental identities, we found that the expression evaluates to 0 for $0^\circ < \theta < 90^\circ$. This detailed process helps in understanding how to manipulate trigonometric expressions effectively.

Revision Table: Key Trigonometric Identities Used

Additional Information: Domain and Range Considerations

The condition $0^\circ < \theta < 90^\circ$ is important. In this interval:

• $\sin\theta$ is between 0 and 1 ($\sin\theta \in (0, 1)$). This means $\sin\theta \ne 0$ and $\sin\theta \ne 1$, ensuring $\operatorname{cosec}\theta$ is defined and $1-\sin\theta \ne 0$.
• $\cos\theta$ is between 0 and 1 ($\cos\theta \in (0, 1)$). This means $\cos\theta \ne 0$, ensuring $\sec\theta$ and $\tan\theta$ are defined.
• $\sin\theta - 1$ is between -1 and 0, so $\sin\theta - 1 \ne 0$, ensuring the denominator in the second term simplification is non-zero.
• All trigonometric functions involved ($\sin, \cos, \tan, \sec, \operatorname{cosec}$) are positive in the first quadrant ($0^\circ < \theta < 90^\circ$).

These conditions ensure that the steps involving division by trigonometric functions or terms like $1 - \sin\theta$ or $\sin\theta - 1$ are valid.