Question : If $\operatorname{cosec} \theta+\cot \theta=\mathbf{s}$, then what is the value of $\cos \theta$?
Option 1: $\frac{s^2-1}{s^2+1}$
Option 2: $\frac{s^2+1}{s^2-1}$
Option 3: $\frac{s^2-1}{2 s}$
Option 4: $\frac{2 s}{s^2+1}$
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Correct Answer: $\frac{s^2-1}{s^2+1}$
Solution :
Given, $\operatorname{cosec} \theta+\cot \theta=\mathbf{s}$
We know, $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin\theta}$
⇒ $\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=s$
⇒ $\frac{1+\cos\theta}{\sin\theta}=s$
Squaring both sides, we get
⇒ $\frac{(1+\cos\theta)^2}{\sin^2\theta}=s^2$
We know, $\cos^2\theta+\sin^2\theta=1$
⇒ $\frac{1+\cos^2\theta+2\cos\theta}{1-\cos^2\theta}=s^2$
Applying componendo and dividendo,
⇒ $\frac{1+\cos^2\theta+2\cos\theta+1-\cos^2\theta}{1+\cos^2\theta+2\cos\theta-(1-\cos^2\theta)}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{1+\cos^2\theta+2\cos\theta+1-\cos^2\theta}{1+\cos^2\theta+2\cos\theta-1+\cos^2\theta}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{2+2\cos\theta}{2\cos^2\theta+2\cos\theta}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{2(1+\cos\theta)}{2\cos\theta(\cos\theta+1)}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{1}{\cos\theta}=\frac{s^2+1}{s^2-1}$
⇒ $\cos\theta=\frac{s^2-1}{s^2+1}$
Hence, the correct answer is $\frac{s^2-1}{s^2+1}$.
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