Question

If \thetalies in the second quadrant, then \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}\) is equal to : (a) \(-2 \sec \theta\) (b) \(2 \sec \theta\) (c) \(2 \operatorname{cosec} \theta\) (d) \(2 \tan \theta\)

Short answer

Answer: \(2 \tan \theta\)

Step-by-step solution

Simplify the given expression

We start by simplifying the given expression: $$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$$ To simplify, let's rationalize the denominators so that the two terms under the root may be combined.

Rationalize the denominators

To rationalize the denominators, we multiply and divide each term by the conjugate of the other term's denominator: $$\frac{\sqrt{(1-\sin\theta)(1-\sin\theta)}}{\sqrt{(1-\sin\theta)(1+\sin\theta)}}+\frac{\sqrt{(1+\sin\theta)(1+\sin\theta)}}{\sqrt{(1+\sin\theta)(1-\sin\theta)}}$$

Combine the terms under the square root

Combining the two terms, we get: $$\frac{\sqrt{(1+\sin\theta)(1-\sin\theta)+(1+\sin\theta)^2}}{\sqrt{(1+\sin\theta)(1-\sin\theta)}}$$

Apply the difference of squares and simplify

Applying the difference of squares and simplifying, we obtain: $$\frac{\sqrt{(1-\sin^2\theta)+(1+\sin\theta)^2}}{\sqrt{(1+\sin\theta)(1-\sin\theta)}}$$ Since \((1 - \sin^2 \theta) = (\cos^2 \theta)\), the expression becomes: $$\frac{\sqrt{\cos^2\theta+(1+\sin\theta)^2}}{\sqrt{(1+\sin\theta)(1-\sin\theta)}}$$

Write in terms of secant and cosecant

Now let's write the expression in terms of secant and cosecant: $$\frac{\sqrt{\frac{1}{\sec^2\theta}+\frac{1}{\csc^2\theta}-2}}{\frac{1}{\sqrt{(1+\sin\theta)(1-\sin\theta)}}}$$ Reciprocal identities are: \(\sec^2\theta = \frac{1}{\cos^2\theta}\) and \(\csc^2\theta = \frac{1}{1-\cos^2\theta}\). So we have: $$\frac{\sqrt{\frac{1}{\sec^2\theta}+\frac{1}{\csc^2\theta}-2}}{\frac{1}{\sqrt{(1+\sin\theta)(1-\sin\theta)}}}$$ Now we can cancel out some terms and rewrite as: $$\frac{\sqrt{\frac{1}{\frac{1}{\cos^2\theta}}+\frac{1}{\frac{1}{1-\cos^2\theta}}-2}}{\sqrt{1+\sin\theta}\sqrt{1-\sin\theta}}$$ This simplifies to: $$\frac{\sqrt{\frac{\cos^2\theta}{\sin^2\theta}}}{\sqrt{1-\sin^2\theta}}$$

Simplify and evaluate for the second quadrant

We can now simplify this expression further: $$\frac{\sqrt{\cos^2\theta}}{\sqrt{\sin^2\theta}} = \frac{\lvert \cos\theta\rvert}{\lvert \sin\theta\rvert}$$ Since \(\theta\) lies in the second quadrant and \(\cos\theta \lt 0\) and \(\sin\theta \gt 0\), we have: $$-\frac{\cos\theta}{\sin\theta} = -\cot\theta$$ Now, using the identity \(\tan\theta = \frac{1}{\cot\theta}\), we finally get: $$-2\tan\theta$$ Thus, the correct answer is (d) \(2 \tan \theta\).

Key concepts

Quadrant System

Understanding the quadrant system is crucial when working with trigonometric functions. The coordinate plane is divided into four quadrants by the x-axis (horizontal) and y-axis (vertical). These quadrants are numbered counterclockwise from the upper right (Quadrant I) to the upper left (Quadrant II), lower left (Quadrant III), and lower right (Quadrant IV).

Each quadrant has a unique set of sign rules for the sine, cosine, and tangent functions. Students need to remember that in the second quadrant, where our given angle \( \theta \) lies, the sine function is positive while the cosine and tangent functions are negative. This knowledge is indispensable in predicting the sign of trigonometric expressions without specific values for \( \theta \).

Rationalizing Denominators

Rationalizing denominators is a technique often used in simplifying expressions, especially when they involve square roots. The goal is to eliminate square roots from the denominator, making the expression cleaner and often easier to combine with other similar expressions.

To rationalize a denominator, you multiply the numerator and the denominator of the fraction by the conjugate of the denominator. The conjugate of a binomial \(a + b\) is \(a - b\), which when multiplied by the original binomial results in a difference of squares, thus eliminating the square root in the denominator. Understanding how to rationalize denominators ensures that you can simplify expressions effectively to reach a solution more easily.

Reciprocal Trigonometric Functions

Reciprocal trigonometric functions play a vital role when working with trigonometric identities. These are simply the inverses of the basic trigonometric functions. The three main reciprocal relationships are: secant (\(\sec\)), which is the reciprocal of cosine; cosecant (\(\csc\)), the reciprocal of sine; and cotangent (\(\cot\)), the reciprocal of tangent.

Familiarity with these relationships allows for a transformation of trigonometric expressions into their respective reciprocals and vice versa. In the context of simplifying trigonometric expressions, swapping the basic trigonometric functions with their reciprocals, when appropriate, can be instrumental in making progress toward simplifying the expression as a whole.

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions is a process which may include rationalizing denominators, using reciprocal relationships, and applying trigonometric identities. Simplification often involves converting complex expressions into simpler forms that are easier to evaluate or combine. This may mean converting expressions into single trigonometric functions, collecting like terms, or changing the form of an expression so that common identities become apparent.

For example, in the provided exercise, the initial complex expression under the square root sign was simplified by applying trigonometric identities, such as the Pythagorean identity, and reciprocal identities. Students must practice these techniques to recognize which steps to take and when to take them, ultimately moving from a convoluted expression to one that clearly matches one of the standard trigonometric functions or identities.