Question

Given $tan\Theta=\frac{1}{√15}$, what is the value of $\frac{cosec^{2}\Theta-sec^{2}\Theta}{cosec^{2}\Theta+sec^{2}\Theta}$

Answer 1

SOLUTION :

Given : tan θ = 1/√5  

In right angle ∆,  

tan θ = perpendicular / Base  = 1/√5

Hypotenuse = √( perpendicular)² + (Base)²

[By Pythagoras theorem]

Hypotenuse = √ 1² + √5² = √1 + 5 = √6

Hypotenuse = √6

cosec θ = Hypotenuse/perpendicular = √6/1

cosec θ = √6

sec θ = Hypotenuse / base = √6/√5

sec θ =  √6/√5

The value of : cosec²θ - sec²θ / cosec²θ + sec²θ

= [ √6² - (√6/√5)² ] / [√6² + (√6/√5)² ]

= [ 6 - 6/5 ] / [ 6 + 6/5 ]  

= [(6 ×5 - 6)/5 ] /  [(6 ×5 + 6)/5 ]

= [(30 - 6)/5] /  [(30 + 6)/5]

= 24/5 / 36/5  

= 24/5 × 5/36

= 24/36 = ⅔  

cosec²θ - sec²θ / cosec²θ + sec²θ = 2/3

Hence, the value of cosec²θ - sec²θ / cosec²θ + sec²θ is ⅔ .  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Solution:

Given that

$tan \:\Theta = \frac{1}{ \sqrt{15} }$
We know that

$1 + {tan}^{2} \Theta = {sec}^{2} \Theta \\ \\ 1 + {cot}^{2} \Theta = {cosec}^{2} \Theta$

apply these identities in the expression to be solved

$\frac{cosec^{2}\Theta-sec^{2}\Theta}{cosec^{2}\Theta+sec^{2}\Theta} \\ \\ \frac{1 + {cot}^{2}\Theta - (1 + {tan}^{2} \Theta) }{1 + {cot}^{2}\Theta + (1 + {tan}^{2} \Theta)} \\ \\ = \frac{ {cot}^{2}\Theta- {tan}^{2}\Theta }{ 2 + {cot}^{2}\Theta + {tan}^{2}\Theta} \\ \\ = \frac{15 - \frac{1}{15} }{2 + 15 + \frac{1}{15} } \\ \\ = \frac{225 - 1}{255 + 1} \\ \\ = \frac{224}{256} \\ \\ \frac{cosec^{2}\Theta-sec^{2}\Theta}{cosec^{2}\Theta+sec^{2}\Theta} = \frac{7}{8}$