Question
If $\text{cosec}\theta+\cot\theta=\text{p},$ then prove that $\cos\theta=\frac{\text{p}^2-1}{\text{p}^2+1}.$
✓
Answer
$\text{cosec}\theta+\cot\theta=\text{p}\ [\text{Given}]$
$\Rightarrow\ \frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\text{p}$
$\Rightarrow\ \frac{1+\cos\theta}{\sin\theta}=\text{p}$
$\Rightarrow\ \frac{(1+\cos\theta)^2}{\sin^2\theta}=\text{p}^2$
$\Rightarrow\ \frac{(1+\cos\theta)^2}{(1-\cos^2\theta)}=\text{p}^2$
$\Rightarrow\ \frac{(1+\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}=\text{p}^2$
$\Rightarrow\ \frac{(1+\cos\theta)}{(1-\cos\theta)}=\frac{\text{p}^2}{1}$
$\begin{bmatrix}\text{By using the rule of componendo and dividendo}\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}\\\text{can be written as }\frac{\text{a}+\text{b}}{\text{a}-\text{b}}=\frac{\text{c}+\text{d}}{\text{c}-\text{d}} \end{bmatrix}$
So,by using componendo and dividendo, we have
$\frac{(1+\cos\theta)+(1-\cos\theta)}{(1+\cos\theta)-(1-\cos\theta)}=\frac{\text{p}^2+1}{\text{p}^2-1}$
$\Rightarrow\ \frac{2}{2\cos\theta}=\frac{\text{p}^2+1}{\text{p}^2-1}\ \Big(\text{By invertendo }\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}};\ \frac{\text{b}}{\text{a}}=\frac{\text{d}}{\text{c}}\Big)$
$\Rightarrow\ \cos\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
Hence, proved.
$\Rightarrow\ \frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\text{p}$
$\Rightarrow\ \frac{1+\cos\theta}{\sin\theta}=\text{p}$
$\Rightarrow\ \frac{(1+\cos\theta)^2}{\sin^2\theta}=\text{p}^2$
$\Rightarrow\ \frac{(1+\cos\theta)^2}{(1-\cos^2\theta)}=\text{p}^2$
$\Rightarrow\ \frac{(1+\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}=\text{p}^2$
$\Rightarrow\ \frac{(1+\cos\theta)}{(1-\cos\theta)}=\frac{\text{p}^2}{1}$
$\begin{bmatrix}\text{By using the rule of componendo and dividendo}\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}\\\text{can be written as }\frac{\text{a}+\text{b}}{\text{a}-\text{b}}=\frac{\text{c}+\text{d}}{\text{c}-\text{d}} \end{bmatrix}$
So,by using componendo and dividendo, we have
$\frac{(1+\cos\theta)+(1-\cos\theta)}{(1+\cos\theta)-(1-\cos\theta)}=\frac{\text{p}^2+1}{\text{p}^2-1}$
$\Rightarrow\ \frac{2}{2\cos\theta}=\frac{\text{p}^2+1}{\text{p}^2-1}\ \Big(\text{By invertendo }\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}};\ \frac{\text{b}}{\text{a}}=\frac{\text{d}}{\text{c}}\Big)$
$\Rightarrow\ \cos\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
Hence, proved.
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