Question

Evaluate \[ \frac{\cos\theta}{1-\cos\theta} + \frac{\sec\theta}{1+\sec\theta}. \]

• A. \(1+2\tan^2\theta\)
• B. \(\sec^2\theta+\csc^2\theta\)
• C. \(\tan^2\theta+\cot^2\theta\)
• D. \(1+\cot^2\theta\) \bigskip

Hint:

Whenever an expression contains \(\sec\theta\), convert it into \(\frac1{\cos\theta}\) first. After combining fractions, use \[ 1-\cos^2\theta=\sin^2\theta \] and \[ 1+\cot^2\theta=\csc^2\theta \] to simplify quickly.

Answer 1

The Correct Option is D

Concept: To simplify trigonometric expressions involving \(\sec\theta\), first express everything in terms of \(\sin\theta\) and \(\cos\theta\). Then combine fractions and use standard identities: \[ 1-\cos^2\theta=\sin^2\theta, \] \[ 1+\cot^2\theta=\csc^2\theta. \]

Step 1: Simplify the second term. Since \[ \sec\theta=\frac{1}{\cos\theta}, \] we have \[ \frac{\sec\theta}{1+\sec\theta} = \frac{\frac1{\cos\theta}} {1+\frac1{\cos\theta}}. \] Multiplying numerator and denominator by \(\cos\theta\), \[ \frac{\sec\theta}{1+\sec\theta} = \frac{1}{1+\cos\theta}. \] Therefore, \[ \frac{\cos\theta}{1-\cos\theta} + \frac{\sec\theta}{1+\sec\theta} = \frac{\cos\theta}{1-\cos\theta} + \frac{1}{1+\cos\theta}. \]

Step 2: Take the LCM and combine the fractions. \[ = \frac{\cos\theta(1+\cos\theta)+(1-\cos\theta)} {(1-\cos\theta)(1+\cos\theta)}. \] \[ = \frac{\cos\theta+\cos^2\theta+1-\cos\theta} {1-\cos^2\theta}. \] \[ = \frac{1+\cos^2\theta} {\sin^2\theta}. \]

Step 3: Rewrite the numerator. \[ 1+\cos^2\theta = (1-\cos^2\theta)+2\cos^2\theta. \] Hence, \[ \frac{1+\cos^2\theta}{\sin^2\theta} = \frac{\sin^2\theta+2\cos^2\theta} {\sin^2\theta}. \] \[ = 1+2\cot^2\theta. \] Now using \[ \csc^2\theta=1+\cot^2\theta, \] we get \[ 1+2\cot^2\theta = (1+\cot^2\theta)+\cot^2\theta = \csc^2\theta+\cot^2\theta. \] Also, \[ \csc^2\theta = 1+\cot^2\theta. \] Thus, \[ \frac{\cos\theta}{1-\cos\theta} + \frac{\sec\theta}{1+\sec\theta} = 1+\cot^2\theta. \]

Step 4: Write the final answer. \[ \boxed{1+\cot^2\theta} \]