Q.If \( \sin\theta - \cos\theta = \frac{7}{13} \), then the value of \( \sin\theta + \cos\theta \) is — (a) \(\cfrac{13}{17}\) (b) \(\cfrac{17}{13}\) (c) \(\cfrac{13}{7}\) (d) None of the above
Answer: B
Given: \( \sin\theta - \cos\theta = \frac{7}{13} \) Then, \( (\sin\theta - \cos\theta)^2 = \left(\frac{7}{13}\right)^2 \) ⇒ \( \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta = \frac{49}{169} \) ⇒ \( 1 - 2\sin\theta\cos\theta = \frac{49}{169} \) ⇒ \( -2\sin\theta\cos\theta = \frac{49}{169} - 1 \) ⇒ \( 2\sin\theta\cos\theta = 1 - \frac{49}{169} = \frac{120}{169} \) Now, \( 1 + 2\sin\theta\cos\theta = 1 + \frac{120}{169} = \frac{289}{169} \) ⇒ \( \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = \frac{289}{169} \) ⇒ \( (\sin\theta + \cos\theta)^2 = \frac{289}{169} \) ⇒ \( \sin\theta + \cos\theta = \sqrt{\frac{289}{169}} = \frac{17}{13} \) So, the value of \( \sin\theta + \cos\theta \) is \( \frac{17}{13} \).
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