Question

If θ is an acute angle such that $sec^{2}\Theta=3$, then the value of $\frac{tan^{2}\Theta-cosec^{2}\Theta }{tan^{2}\Theta+cosec^{2}\Theta }$ is
(a)$\frac{4}{7}$
(b)$\frac{3}{7}$
(c)$\frac{2}{7}$
(d)$\frac{1}{7}$

Answer 1

SOLUTION :  

The correct option is (d) : 1/ 7

Given : sec² θ  = 3

sec θ = √3/1

In right angle ∆ ,  

sec θ =  Hypotenuse /base = √3/1

Hypotenuse = √3 , base = 1

Hypotenuse² = ( perpendicular)² + (Base)²

[By Pythagoras theorem]

√3²  =  (perpendicular)² + (1)²  

3 =  (perpendicular)² + (1)

(perpendicular)² = 3 - 1

(perpendicular)² = 2  

perpendicular = √2

tan θ =  perpendicular/base = √2/1  

tan θ = √2

cosec θ = Hypotenuse/perpendicular = √3/√2

cosec θ = √3/√2

The value of : tan² θ - cosec² θ / tan² θ + cosec² θ  

= (√2)² - (√3/√2)² / (√2)² + (3/√2)²

= ( 2 - 3/2) / ( 2 + 3/2)

= ( 2/1 - 3/2) / ( 2/1 + 3/2)

= [(2×2 - 3)/2 ] /  [(2×2 + 3)/2 ]  

= [(4 - 3)/2] / [(4 + 3)/2]

= (½) / (7/2)

= ½ × 2/7

= 1/7

tan² θ - cosec² θ / tan² θ + cosec² θ = 1/7

Hence, the value of tan² θ - cosec² θ / tan² θ + cosec² θ is 1/7 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

The correct option is d. 1/7