Question

if 6x=sec theta and 6/x=tan theta, find the value of 9(x²-1/x²)​

Answer 1

Answer:

Given that

6x = sec theta

6/x = tan theta

Let

6x = sec theta ===> eqn 1

6/x = tan theta ===> eqn 2

Add eqn 1 and 2 on both sides

6x + 6/x = sec theta + tan theta

6(x+1/x) = (sec theta+tan theta)=> eqn 3

Now ,

Substrate eqn 1 and 2 on both sides

6x - 6/x = sec theta - tan theta

6(x-1/x) = (sec theta - tan theta) => eqn 4

Multiply eqn 3 and 4 on both sides

eqn 3 =>6(x+1/x)= (sec theta+tan theta)

eqn 4 =>6(x-1/x) = (sec theta - tan theta)

6(x+1/x) . 6(x-1/x) = ( sec theta + tan theta ) . ( sec theta - tan theta )

36 ( x² - 1/x² ) = sec²theta - tan²theta

9×4 ( x² - 1/x² ) = 1

9( x² - 1/x² ) = 1 / 4

Step-by-step explanation:

The Value of 9 ( x² - 1/x² ) is 1/4