Question

If \[ S=\left\{\theta\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right]:\cos^2\theta+\sin\theta\tan\theta=\cos2\theta\right\, \] then \[ \sum_{\theta\in S}(\sin\theta+\cos\theta)= \]

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

Whenever \(\tan\theta\) appears with \(\sin\theta\) and \(\cos\theta\), convert everything into \(\sin\theta\) and \(\cos\theta\). This often reduces the equation to a simple factorization.

Answer 1

The Correct Option is C

Concept: The given equation contains a mixture of trigonometric functions. The first step is to express everything in terms of \(\sin\theta\) and \(\cos\theta\). After simplification, we solve the resulting trigonometric equation in the given interval and then evaluate the required summation. Useful identities: \[ \tan\theta=\frac{\sin\theta}{\cos\theta}, \] \[ \cos2\theta=\cos^2\theta-\sin^2\theta, \] \[ \sin^2\theta+\cos^2\theta=1. \]

Step 1: Simplify the given equation. Given, \[ \cos^2\theta+\sin\theta\tan\theta=\cos2\theta. \] Substituting \[ \tan\theta=\frac{\sin\theta}{\cos\theta}, \] we get \[ \cos^2\theta+\frac{\sin^2\theta}{\cos\theta} = \cos2\theta. \] Using \[ \cos2\theta=\cos^2\theta-\sin^2\theta, \] the equation becomes \[ \cos^2\theta+\frac{\sin^2\theta}{\cos\theta} = \cos^2\theta-\sin^2\theta. \] Cancelling \(\cos^2\theta\) from both sides, \[ \frac{\sin^2\theta}{\cos\theta} = -\sin^2\theta. \]

Step 2: Factor the equation. Multiplying both sides by \(\cos\theta\), \[ \sin^2\theta = -\sin^2\theta\cos\theta. \] \[ \sin^2\theta(1+\cos\theta)=0. \] Therefore, \[ \sin\theta=0 \] or \[ \cos\theta=-1. \]

Step 3: Find all solutions in the interval. The interval is \[ \left[\frac{\pi}{2},\frac{3\pi}{2}\right]. \] For \[ \sin\theta=0, \] the solutions are \[ \theta=\pi. \] For \[ \cos\theta=-1, \] the solution is again \[ \theta=\pi. \] Hence, \[ S=\{\pi\}. \]

Step 4: Evaluate the required sum. \[ \sum_{\theta\in S}(\sin\theta+\cos\theta) = \sin\pi+\cos\pi. \] Since \[ \sin\pi=0, \qquad \cos\pi=-1, \] we obtain \[ 0+(-1)=-1. \] Conclusion: \[ \boxed{-1} \]