Express $tan^{-1}(\frac{cos~x}{1-sin~x})$ where $\frac{-\pi}{2}<x<\frac{\pi}{2}$ in the simplest form.

Q1310: Express $tan^{-1}(\frac{cos~x}{1-sin~x})$ where $\frac{-\pi}...

Express $tan^{-1}(\frac{cos~x}{1-sin~x})$ where $\frac{-\pi}{2}

AI Tutor Explanation

Detailed Solution

Step 1: Simplify the expression inside the inverse tangent function.

We are given the expression $tan^{-1}(\frac{cos~x}{1-sin~x})$. We need to simplify the fraction $\frac{cos~x}{1-sin~x}$ using trigonometric identities.

Step 2: Use trigonometric identities to rewrite cos(x) and sin(x).

We can rewrite $cos~x$ as $cos^2(\frac{x}{2}) - sin^2(\frac{x}{2})$ and $sin~x$ as $2sin(\frac{x}{2})cos(\frac{x}{2})$. So, $\frac{cos~x}{1-sin~x} = \frac{cos^2(\frac{x}{2}) - sin^2(\frac{x}{2})}{1 - 2sin(\frac{x}{2})cos(\frac{x}{2})}$

Step 3: Rewrite the denominator as a perfect square.

The denominator can be written as $(cos^2(\frac{x}{2}) + sin^2(\frac{x}{2})) - 2sin(\frac{x}{2})cos(\frac{x}{2}) = (cos(\frac{x}{2}) - sin(\frac{x}{2}))^2$. The numerator can be factored as $(cos(\frac{x}{2}) - sin(\frac{x}{2}))(cos(\frac{x}{2}) + sin(\frac{x}{2}))$. Therefore, $\frac{cos~x}{1-sin~x} = \frac{(cos(\frac{x}{2}) - sin(\frac{x}{2}))(cos(\frac{x}{2}) + sin(\frac{x}{2}))}{(cos(\frac{x}{2}) - sin(\frac{x}{2}))^2} = \frac{cos(\frac{x}{2}) + sin(\frac{x}{2})}{cos(\frac{x}{2}) - sin(\frac{x}{2})}$

Step 4: Divide both numerator and denominator by $cos(\frac{x}{2})$.

Dividing both numerator and denominator by $cos(\frac{x}{2})$, we get: $\frac{1 + tan(\frac{x}{2})}{1 - tan(\frac{x}{2})}$

Step 5: Recognize the tangent addition formula.

We know that $tan(a+b) = \frac{tan~a + tan~b}{1 - tan~a~tan~b}$. Here, we can write $\frac{1 + tan(\frac{x}{2})}{1 - tan(\frac{x}{2})} = \frac{tan(\frac{\pi}{4}) + tan(\frac{x}{2})}{1 - tan(\frac{\pi}{4})tan(\frac{x}{2})} = tan(\frac{\pi}{4} + \frac{x}{2})$

Step 6: Substitute back into the original expression.

So, $tan^{-1}(\frac{cos~x}{1-sin~x}) = tan^{-1}(tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$

Step 7: Check the range of the simplified expression.

Given that $\frac{-\pi}{2} < x < \frac{\pi}{2}$, we have $\frac{-\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$. Adding $\frac{\pi}{4}$ to all sides, we get $0 < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$. Since the range of $tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, the simplified expression is valid.

Final Answer: $\frac{\pi}{4} + \frac{x}{2}$

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand trigonometric identities and apply them to simplify the given expression. The student must also understand the range of inverse trigonometric functions.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding of trigonometric identities and their application in simplifying expressions, which falls under the conceptual knowledge dimension.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of inverse trigonometric functions and trigonometric identities, which are core concepts in the syllabus.

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