Aspire Faculty ID #18012 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

Let $\frac{\pi}{2} < \theta < \pi$ and $\cot\theta = -\frac{1}{2\sqrt{2}}$. Then the value of

$\sin\left(\frac{150}{2}\right)(\cos80 + \sin80) + \cos\left(\frac{150}{2}\right)(\cos80 - \sin80)$

is equal to:

Solution

$\frac{\pi}{2} < \theta < \pi,; \cot\theta = -\frac{1}{2\sqrt{2}}$

$\Rightarrow \sin\left(\frac{150}{2}\right)(\cos80 + \sin80) + \cos\left(\frac{150}{2}\right)(\cos80 - \sin80)$

$= \sin\left(\frac{150}{2}\right)\cos80 - \cos\left(\frac{150}{2}\right)\sin80 + \sin\left(\frac{150}{2}\right)\sin80 + \cos\left(\frac{150}{2}\right)\cos80$

$= \sin\left(\frac{150}{2} - 80\right) + \cos\left(\frac{150}{2} - 80\right)$

$= \cos\left(\frac{\theta}{2}\right) - \sin\left(\frac{\theta}{2}\right)$

$= \sqrt{1 - \sin\theta}$

Given $\cot\theta = \frac{\cos\theta}{\sin\theta} = -\frac{1}{2\sqrt{2}}$

$\Rightarrow \sin\theta = \frac{2\sqrt{2}}{3}$

$\Rightarrow \sqrt{1 - \sin\theta} = \sqrt{\frac{3 - 2\sqrt{2}}{3}} = \frac{\sqrt{2} - 1}{\sqrt{3}}$

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