Question

If x = a sec θ and y = b tan θ, then b²x²− a²y² =
A. ab
B. a²− b²
C. a² + b²
D. a² b²

Answer 1

b²x²-a²x² = a²b²

•x = a secθ

•y = b tanθ

•x² = a² sec²θ

•y² = b² tan²θ

•b²x² = a²b² sec²θ

•a²y² = a²b² tan²θ

•b²x²-a²x² = a²b² sec²θ - a²b² tan²θ

= a²b² ( sec²θ - tan²θ)

•As, sec²θ = 1 + tan²θ

•sec²θ - tan²θ =1

•b²x²-a²x² = a²b² (1)

•b²x²-a²x² = a²b²

Answer 2

Option D: $a^{2} b^{2}$ is the value of the expression $b^{2} x^{2}-a^{2} y^{2}$

Explanation:

Given that $x=a \sec \theta$ and $y=b \tan \theta$

We need to determine the value of the expression $b^{2} x^{2}-a^{2} y^{2}$

Let us substitute the values $x=a \sec \theta$ and $y=b \tan \theta$ in the expression $b^{2} x^{2}-a^{2} y^{2}$, we get,

$b^{2} x^{2}-a^{2} y^{2}=b^2(a \ sec\ \theta)^2-a^2({b} \tan \theta)^2$

Simplifying, we get,

$b^{2} x^{2}-a^{2} y^{2}=b^2a^2 \ sec^2\ \theta-a^2{b^2} \tan^2 \theta$

Taking out the common terms, we have,

$b^{2} x^{2}-a^{2} y^{2}=a^2 b^2(\ sec^2\ \theta-\tan^2 \theta)$

Since, we know the identity, $1+\tan ^{2} \theta=\sec ^{2} \theta$

Thus, we have,

$b^{2} x^{2}-a^{2} y^{2}=a^2 b^2(1)$

Simplifying, we get,

$b^{2} x^{2}-a^{2} y^{2}=a^2 b^2$

Thus, the value of the expression is $a^{2} b^{2}$

Therefore, Option D is the correct answer.

Learn more:

(1) If x = a sec θ and y = b tan θ, thenb²x²-a²y² =

(a)ab

(b)a² − b²

(c)a² + b²

(d)a² b²

brainly.in/question/11828258

(2) Prove the identity : (sec² θ -1) (cosec² θ -1) = 1

brainly.in/question/3779224