Question

if x cosec theta=r, y sec theta=p and r sin^2theta+p cos^2theta=z then which of the following relations is correct​

Answer 1

$\textbf{Given:}$

$x\;cosec\theta=r,\;\;y\;sec\theta=p$

$\implies\;cosec\theta=\frac{r}{x},\;\;sec\theta=\frac{p}{y}$

$\text{Taking reciprocals on bothsides, we get}$

$sin\theta=\frac{x}{r},\;\;cos\theta=\frac{y}{p}$

$\text{Now,}$

$r\;sin^2\theta+p\;cos^2\theta=z$

$\impliesr(\frac{x^2}{r^2})+p(\frac{y^2}{p^2})=z$

$\implies\boxed{\bf\frac{x^2}{r}+\frac{y^2}{p}=z}$

$\text{which is the required relation}$

Answer 2

Given:

$\bold{ r =x \ cosec \theta}\\\\\bold{p= y \sec \theta}\\\\\bold{z= r\sin^2 \theta +p \cos^2 \theta}$

To prove:

Find relation=?

Solution:

In the given question there is no relation choices were given. so, the relation to this question can be defined as follows:

$\Rightarrow$ $z= r\sin^2 \theta +p \cos^2 \theta$

put the value of r and z in the above equation:

$\Rightarrow z= (x \ cosec \theta)\sin^2 \theta +(y \sec \theta) \cos^2 \theta\\\\\Rightarrow z= (x \times \frac{1}{\sin \theta})\times \sin^2 \theta +(y \times \frac{1}{\cos \theta})\times \cos^2 \theta\\\\\Rightarrow z= x \times \frac{1}{\sin \theta}\times \sin^2 \theta +y \times \frac{1}{\cos \theta}\times \cos^2 \theta\\\\\Rightarrow \boxed{z= x \sin \theta +y \cos \theta}$

                                                             OR

$\to r =x \ cosec \theta....(i)\\\\ \to p= y \sec \theta.... (ii) \\\\ \to z= r\sin^2 \theta +p \cos^2 \theta....(iii)$

Solve the equation (i), (ii) and put the value in equation (iii) :

$\to r =x \frac{1}{\sin \theta}\\\\ \to \sin \theta = \frac{x}{r} \\\\ \to p= y \frac{1}{\cos \theta}\\\\ \to \cos \theta = \frac{y}{p}$

Equation (iii):

$\to z= r\sin^2 \theta +p \cos^2 \theta\\\\\to z= r\times (\frac{x}{r})^2 +p \times (\frac{y}{p})^2\\\\\to z= r\times \frac{x^2}{r^2} +p \times \frac{y^2}{p^2}\\\\\to \boxed{\boxed{z= \frac{x^2}{r} +\frac{y^2}{p}}}$