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प्रश्न
Find x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]
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उत्तर
\[90^\circ = \frac{\pi}{2}\]
We have:
\[ x \cot\left( 90^\circ + \theta \right) + \tan\left( 90^\circ + \theta \right) \sin \theta + cosec\left( 90^\circ + \theta \right) = 0\]
\[ \Rightarrow x \left[ - \tan \theta \right] + \left[ - \cot \theta \right] \sin \theta + \sec \theta = 0\]
\[ \Rightarrow - x \tan \theta - \cot \theta \sin \theta + \sec \theta = 0 \]
\[ \Rightarrow - x \times \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} \times \sin \theta + \frac{1}{\cos \theta} = 0 \]
\[ \Rightarrow - x \times \frac{\sin \theta}{\cos \theta} - \cos\theta + \frac{1}{\cos \theta} = 0 \]
\[ \Rightarrow \frac{- x \sin \theta - \cos^2 \theta + 1}{\cos \theta} = 0 \]
`=>-xsintheta-cos^2theta+1=0`
`=>-xsintheta = - sin^2theta ...[∵ 1 - cos^2theta = sin^2theta]`
`=>x=(-sin^2theta)/(-sintheta)`
`=>x=sintheta`
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