Question

if maximum and minimum values of 7+ 6tan theta - tan^2 theta /(1+tan^2 theta) i for all real values theta ( theta not equal π/2) are lambda and nu respectively. then lambda- nu​

Answer 1

Given :

• $\frac{7+6tan\theta-tan^{2}\theta}{1+tan^{2}\theta}$

• maximum = $\lambda$

• minimum = $\nu$

To find : $\lambda -\nu$

Solution :

• Refer to the added picture to see the graph of the function.

• You can see clearly that the height above the x -axis is the maximum, i.e., 8.

• And the depth below the x -axis is the minimum, i.e., (- 2).

• Then, $\lambda=8$ and $\nu=-2$

• So, $\lambda-\nu=8-(-2)=8+2=10$

Answer : $\lambda-\nu=10$

Answer 2

Answer:

Step-by-step explanation:

ANSWER IS BELOW