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Trigonometric Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\text{x}=\text{r}\sin\theta\cos\theta,\text{y}=\text{r}\sin\theta$ and $\text{z}=\text{r}\cos\theta,$ then $\text{x}^2+\text{x}^2+\text{z}^2$ is idepandent of
  • ✓
    $\theta,\phi$
  • B
    $\text{r},\theta$
  • C
    $\text{r},\phi$
  • D
    $\text{r}.$
✓

Answer

Correct option: A.
$\theta,\phi$
We have :
$\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\theta\sin\phi$  and $\text{z}=\text{r}\cos\theta,$
$\therefore\text{x}^2+\text{y}^2+\text{z}^2$
$=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$
$=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$
$=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$
$=\text{r}^2\sin^2\theta\times1+\text{r}^2\cos^2\theta$
$=\text{r}^2\sin^2\theta+\text{r}\cos^2\theta$
$=\text{r}^2(\sin^2\theta+\cos^2\theta)$
$=\text{r}^2\times1$
$=\text{r}^2$
Thus, $\text{x}^2+\text{y}^2+\text{z}^2$ is independent of $\theta$ and $\phi$

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