Question

If x cosθ =a and y = a tanθ, then prove that x2–y2=a2

Answer 1

Step-by-step explanation:

Given,

$x \cos \theta = a$

and

$y = a \tan \theta$

To prove

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

Formula to be used

$\cos \theta = \frac{1}{ \sec \theta} \\ \implies \frac{1}{ \cos \theta} = \sec \theta$

and

$\sec ^{2} \theta = {tan}^{2} \theta + 1 \\ \implies \sec^{2} \theta - \tan^{2} \theta = 1$

Now from the given data we have

$x \cos \theta = a \\ \implies x = \frac{a}{ \cos \theta} \\ \implies x = a \sec \theta \\ \implies x ^{2} = {a}^{2} \sec ^{2} \theta$

and

$y = a \tan \theta \\ \implies {y}^{2} = {a}^{2} \tan ^{2} \theta$

Now LHS

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

$Hence \: proved$

Answer 2

Solutions:

Given

x cos(theta)=a

and

y = S tan(theta)

x² - y² = a²

cos(theta) = 1/sec(theta)

1/cos(theta) = sec(theta)

sec²(theta)=tan²(theta) + 1

sec²(theta)-tan²(theta) = 1

x cos(theta) = a

x = a/cos(theta)

x = a sec(theta)

x² = a² sec²(theta)

y =a tan(theta)

y²= a tan2(theta)

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