MCQ
If $\text{x}=\text{a}\sec\theta$ and ${y}=\text{b}\tan\theta,$ then $b^2x^2- a^2y^2=$
- A$ab$
- B$a^2- b^2$
- C$a^2+ b^2$
- ✓$a^2b^2$
✓
Answer
Correct option: D.
$a^2b^2$
Given, $\text{x}=\text{a}\sec\theta,\text{y}=\text{b}\tan\theta$
So,
$\text{b}^2\text{y}^2-\text{a}^2\text{y}^2$
$=\text{b}^2(\text{a}\sec\theta)^2-\text{a}^2(\text{b}\tan\theta)^2$
$=\text{b}^2\text{a}^2\sec^2\theta-\text{a}^2\text{b}^2\tan^2\theta$
$=\text{b}^2\text{a}^2(\sec^2\theta-\tan^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\text{b}^2\text{x}^2-\text{a}^2\text{y}^2=\text{a}^2\text{b}^2$
Hence, the correct option is $(D).$
So,
$\text{b}^2\text{y}^2-\text{a}^2\text{y}^2$
$=\text{b}^2(\text{a}\sec\theta)^2-\text{a}^2(\text{b}\tan\theta)^2$
$=\text{b}^2\text{a}^2\sec^2\theta-\text{a}^2\text{b}^2\tan^2\theta$
$=\text{b}^2\text{a}^2(\sec^2\theta-\tan^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\text{b}^2\text{x}^2-\text{a}^2\text{y}^2=\text{a}^2\text{b}^2$
Hence, the correct option is $(D).$
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