Question
If $\text{cosec }\theta+\cot\theta=\text{p},$ prove that $\cos\theta=\frac{\big(\text{p}^2-1\big)}{\big(\text{p}^2+1\big)}.$
✓
Answer
$\text{cosec}\theta+\cot\theta=\text{p}$$\Rightarrow\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\text{p}$
$\Rightarrow\frac{1+\cos\theta}{\sin\theta}=\text{p}$
Squaring both sides, we get:
$\Big(\frac{1+\cos\theta}{\sin\theta}\Big)^2=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{\sin^2\theta}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{1-\cos^2\theta}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)}{(1-\cos\theta)}=\text{p}^2$
$\Rightarrow1+\cos\theta=\text{p}^2(1-\cos\theta)$
$\Rightarrow1+\cos\theta=\text{p}^2-\text{p}^2\cos\theta$
$\Rightarrow\cos\theta\big(1+\text{p}^2\big)=\text{p}^2-1$
$\Rightarrow\cos\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
Hence proved.
$\Rightarrow\frac{1+\cos\theta}{\sin\theta}=\text{p}$
Squaring both sides, we get:
$\Big(\frac{1+\cos\theta}{\sin\theta}\Big)^2=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{\sin^2\theta}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{1-\cos^2\theta}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)}{(1-\cos\theta)}=\text{p}^2$
$\Rightarrow1+\cos\theta=\text{p}^2(1-\cos\theta)$
$\Rightarrow1+\cos\theta=\text{p}^2-\text{p}^2\cos\theta$
$\Rightarrow\cos\theta\big(1+\text{p}^2\big)=\text{p}^2-1$
$\Rightarrow\cos\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
Hence proved.
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