यदि किसी x in (pi, frac{3pi}{2}) के लिए cot x | vedclass

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(C) दिया है $\cot x = \frac{5}{12}$ और $x \in (\pi, \frac{3\pi}{2})$, अतः $\cos x = -\frac{5}{13}$ और $\sin x = -\frac{12}{13}$.चूँकि $\frac{x}{2} \in

Vedclass · Accepted Answer

$\frac{1}{\sqrt{13}}$

Vedclass · Answer

(C) दिया है $\cot x = \frac{5}{12}$ और $x \in (\pi, \frac{3\pi}{2})$, अतः $\cos x = -\frac{5}{13}$ और $\sin x = -\frac{12}{13}$.चूँकि $\frac{x}{2} \in (\frac{\pi}{2}, \frac{3\pi}{4})$, $\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \frac{3}{\sqrt{13}}$ और $\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\frac{2}{\sqrt{13}}$.व्यंजक का सरलीकरण: $\cos(7x - \frac{13x}{2}) + \sin(7x - \frac{13x}{2}) = \cos \frac{x}{2} + \sin \frac{x}{2}$.मान रखने पर: $-\frac{2}{\sqrt{13}} + \frac{3}{\sqrt{13}} = \frac{1}{\sqrt{13}}$.

यदि किसी $x \in (\pi, \frac{3\pi}{2})$ के लिए $\cot x = \frac{5}{12}$ है, तो $\sin 7x(\cos \frac{13x}{2} + \sin \frac{13x}{2}) + \cos 7x(\cos \frac{13x}{2} - \sin \frac{13x}{2})$ का मान ज्ञात कीजिए।

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