Question

If $cos\Theta=\frac{2}{3}$, then $2sec^{2}\Theta+2tan^{2}\Theta-7$ is equal to
(a)1
(b)0
(c)3
(d)4

Answer 1

SOLUTION :  

The correct option is (b) : 0  

Given : cos θ = ⅔

cos θ = base/ hypotenuse = ⅔

Base = 2, hypotenuse = 3

In right angle ∆ ,  

Hypotenuse² = (perpendicular)² + (Base)²

[By Pythagoras theorem]

3² = (perpendicular)² + 2²

9 = (perpendicular)² + 4

perpendicular² = 9 - 4 = 5  

perpendicular = √5

tan θ =  perpendicular/base = √5/2

tan θ = √5/2

sec θ = 1/cos θ = 1/(2/3) = 3/2  

sec θ = 3/2

The value of : 2 sec² θ + 2 tan² θ -7  

= 2(3/2)² + 2(√5/2)² - 7

= 2 × 9/4 + 2 × 5/4 - 7

= (9/2 + 5/2) - 7

= (9 + 5)/2 - 7

= 14/2 - 7  

= 7 - 7

= 0  

2 sec² θ + 2 tan² θ -7 = 0

Hence, the value of 2 sec² θ + 2 tan² θ -7 is 0.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

Option(B)

Step-by-step explanation:

Given, cosθ = (2/3).

We know that sinθ = √1 - (2/3)²

                               = √1 - 4/9

                               = √5/9

                               = (√5)/(3)

Then:

2 sec²θ + 2 tan² θ - 7

= 2 (1/cosθ)² + 2(sinθ/cosθ)² - 7

= 2(3/2)² + 2(√5/2)² - 7

= 2(9/4) + 2(5/4) - 7

= (9/2) + (5/2) - 7

= (-14 + 9 + 5)/2

= 0.

Hope it helps!