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Trigonometric Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions2 Marks
Question
Find x from the following equation: $\text{coses}\Big(\frac{\pi}{2}+\theta\Big)+\text{x}\cos\theta\cot\Big(\frac{\pi}{2}+\theta\Big)=\sin\Big(\frac{\pi}{2}+\theta\Big)$
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Answer

$\text{coses}(90^\circ+\theta)+\text{x}\cos\theta\cot(90^\circ+\theta)=\sin(90^\circ+\theta)$ $\Rightarrow \sec+\text{x}\cos\theta\times(-\tan\theta)=\cos\theta$ $\Rightarrow\frac{1}{\cos\theta}+\text{x}\cos\theta\times\frac{(-\sin\theta)}{\cos\theta}=\cos\theta$ $\Rightarrow\frac{1}{\cos\theta}-\text{x}\sin\theta=\cos\theta$ $\Rightarrow\frac{1-\text{x}\sin\theta\cos\theta}{\cos\theta}=\cos\theta$ $\Rightarrow1-\text{x}\sin\theta\cos\theta=\cos^2\theta$ $\Rightarrow1-\cos^2\theta=\text{x}\sin\theta\cos\theta$ $\Rightarrow\sin^2\theta=\text{x}\sin\theta\cos\theta$ $\Rightarrow\sin\theta=\text{x}\cos\theta$ $\Rightarrow\text{x}=\frac{\sin\theta}{\cos\theta}$ $=\tan\theta$ $\text{Hence x}=\tan\theta$

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