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Find X from the Following Equations: C O S E C ( π 2 + θ ) + X Cos θ Cot ( π 2 + θ ) = Sin ( π 2 + θ ) - Mathematics

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Question

Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]

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Solution

\[90^\circ = \frac{\pi}{2}\]
 We have: 
\[ cosec\left( 90^\circ + \theta \right) + x \cos \theta \cot\left( 90^\circ + \theta \right) = \sin\left( 90^\circ + \theta \right)\]
\[ \Rightarrow \sec \theta + x \cos \theta \left[ - \tan \theta \right] = \cos \theta\]
\[ \Rightarrow \sec \theta - x cos\theta tan\theta = \cos \theta\]
\[ \Rightarrow \sec \theta - x cos\theta \times \frac{\sin \theta}{\cos \theta} = \cos \theta\]
\[ \Rightarrow \sec \theta - x \sin\theta = \cos \theta\]
\[ \Rightarrow \sec \theta - \cos \theta = x \sin\theta$\]
\[ \Rightarrow \frac{1}{\cos \theta} - cos\theta = x \sin\theta\]
\[ \Rightarrow \frac{1 - \cos^2 \theta}{\cos \theta} = x \sin\theta$\]
\[ \Rightarrow \frac{\sin^2 \theta}{cos\theta} = x \sin\theta\]
\[ \Rightarrow \frac{\sin^2 \theta}{\cos \theta \sin \theta} = x\]
\[ \Rightarrow \frac{\sin \theta}{\cos \theta} = x\]
\[ \Rightarrow \tan\theta = x\]
\[ \therefore x = \tan\theta\]

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Chapter 5: Trigonometric Functions - Exercise 5.3 [Page 40]

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RD Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 8.1 | Page 40

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