Question

The locus of the point (sec theta + tan theta),(sec theta- tan theta)

Answer 1

Given:

(sec theta + tan theta),(sec theta- tan theta)

To find:

The locus of the point (sec theta + tan theta),(sec theta- tan theta)

Solution:

From given, we have,

(sec theta + tan theta),(sec theta- tan theta)

Let,

h = (sec theta + tan theta) and k = (sec theta- tan theta)

multiply both the equations, we get,

h × k = (sec theta + tan theta) × (sec theta- tan theta)

h × k = sec² theta -  sec theta × tan theta + sec theta × tan theta - tan² theta

h × k = sec² theta - tan² theta

using the trigonometric functions property, we have,

h × k = 1

⇒ x × y = 1

Therefore, the locus of the point (sec theta + tan theta),(sec theta- tan theta) is xy = 1.

Answer 2

Answer:

Step-by-step explanation: