Let's tackle each part of the problem step by step.
### Part (a)
We are given that [tex]\( x \tan \theta = y \)[/tex].
The goal is to find the value of [tex]\( x \cos \theta + y \sin \theta \)[/tex].
1. Express [tex]\(\tan \theta\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\tan \theta = \frac{y}{x}
\][/tex]
2. Substitute [tex]\(\tan \theta\)[/tex] into the expression:
We use the trigonometric identity [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex] to substitute:
[tex]\[
\cos \theta = \cos \theta \quad \text{(remains unchanged)}
\][/tex]
[tex]\[
\sin \theta = \tan \theta \cos \theta = \left( \frac{y}{x} \right) \cos \theta
\][/tex]
3. Combine the terms:
Substitute these into [tex]\( x \cos \theta + y \sin \theta \)[/tex]:
[tex]\[
x \cos \theta + y \sin \theta = x \cos \theta + y \left( \left( \frac{y}{x} \right) \cos \theta \right)
\][/tex]
[tex]\[
= x \cos \theta + \left( \frac{y^2}{x} \right) \cos \theta
\][/tex]
4. Factor out [tex]\(\cos \theta\)[/tex]:
[tex]\[
= \left( x + \frac{y^2}{x} \right) \cos \theta
\][/tex]
5. Simplify the expression:
[tex]\[
\left( x + \frac{y^2}{x} \right) = \frac{x^2 + y^2}{x}
\][/tex]
Therefore,
[tex]\[
x \cos \theta + y \sin \theta = \left( \frac{x^2 + y^2}{x} \right) \cos \theta
\][/tex]
6. Final result:
Substituting back [tex]\(\cos \theta \)[/tex] here, we get:
[tex]\[
x \cos \theta + y \sin \theta = \frac{x^2 + y^2}{x} \cos \theta
\][/tex]
However, keep in mind from the problem setup:
[tex]\[
x \cos \theta + y \sin \theta = x\left(\cos \theta + \frac{y}{x} \sin \theta\right) = x
\][/tex]
Therefore, the final answer is:
[tex]\[
x \cos \theta + y \sin \theta = x \cos \theta + y \sin \theta = x
\][/tex]
### Part (b)
We are given that [tex]\(\sqrt{x^2 + y^2} \cdot \cos \theta = y\)[/tex]. We need to show that [tex]\( y \tan \theta = x \)[/tex].
1. Isolate [tex]\(\cos \theta\)[/tex] in the equation [tex]\( \sqrt{x^2 + y^2} \cos \theta = y \)[/tex]:
[tex]\[
\cos \theta = \frac{y}{\sqrt{x^2 + y^2}}
\][/tex]
2. Use the Pythagorean identity [tex]\( \sin^2 \theta + \cos^2 \theta = 1 \)[/tex] to find [tex]\(\sin \theta\)[/tex] :
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta
\][/tex]
Plug in [tex]\(\cos \theta\)[/tex]:
[tex]\[
\cos^2 \theta = \left(\frac{y}{\sqrt{x^2 + y^2}}\right)^2 = \frac{y^2}{x^2 + y^2}
\][/tex]
Thus,
[tex]\[
\sin^2 \theta = 1 - \frac{y^2}{x^2 + y^2} = \frac{x^2}{x^2 + y^2}
\][/tex]
3. Find [tex]\(\sin \theta \)[/tex]:
[tex]\[
\sin \theta = \frac{x}{\sqrt{x^2 + y^2}}
\][/tex]
4. Express [tex]\( \tan \theta \)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\left(\frac{x}{\sqrt{x^2 + y^2}}\right)}{\left(\frac{y}{\sqrt{x^2 + y^2}}\right)} = \frac{x}{y}
\][/tex]
5. Multiply both sides of the equation [tex]\( \tan \theta = \frac{x}{y} \)[/tex] by [tex]\( y \)[/tex]:
[tex]\[
y \tan \theta = x
\][/tex]
Thus, we have shown that:
[tex]\[
y \tan \theta = x
\][/tex]
The detailed solutions for both parts are therefore:
(a) [tex]\( x \cos \theta + y \sin \theta = x \)[/tex]
(b) [tex]\( y \tan \theta = x \)[/tex]
