Question

If $tan\Theta=\frac{1}{\sqrt{7} }$, then cosec2 θ-sec2 θcosec2 θ+sec2 θ=
(a)$\frac{5}{7}$
(b)$\frac{3}{7}$
(c)$\frac{1}{7}$
(d)$\frac{3}{4}$

Answer 1

SOLUTION :  

The correct option is (d) : 3/4

Given : tan θ  = 1/√7

In right angle ∆ ,  

tan θ =  perpendicular/base = 1/√7

perpendicular = 1 , base = √7

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

[By Pythagoras theorem]

Hypotenuse = √ 1² + √7² = √1 + 7 = √8

Hypotenuse = √8

cosec θ = hypotenuse/ perpendicular = √8/1

cosec θ = √8

sec θ = hypotenuse/ base =  √8/√7

sec θ = √8/√7

The value of : (cosec² θ - sec² θ) / (cosec² θ + sec² θ)

= [(√8)² - (√8/√7)²] / [(√8)² + (√8/√7)²]  

= (8/1 - 8/7) / (8/1 + 8/7)

= [(8×7 - 8)/7] /  [(8×7 + 8)/7]  

= [(56 - 8 )/7 ] /  [(56 + 8 )/7 ]

= (48/7) / (64/7)

= (48/7) ×( 7/64)

= 48/64

= ¾  

(cosec² θ - sec² θ) / (cosec² θ + sec² θ) = 3/4

Hence, the value of  (cosec² θ - sec² θ) / (cosec² θ + sec² θ) is ¾ .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Option ( D ) is correct.

Here I am using A instead of theta.

Given tan A = 1/√7

=> tan² A = 1/7 -----( 1 )

cot² A = 1/tan²A = 7 ----( 2 )

Now ,

Value of

( cosec²A-sec²A)/(cosec² A + sec²A )

= [(1+cot²A)-(1+tan²A)]/[(1+cot²A)+(1+tan²A)]

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By Trigonometric identities :

1 ) sec²A - tan²A = 1

2 ) cosec² A - cot²A = 1

**************************************

= ( 1+cot²A-1-tan²A)/(1+cot²A+1+tan²A)

= ( cot²A - tan²A )/( 2+cot²A+tan²A )

= [ 7 - 1/7 ]/[ 2 + 7 + 1/7 ]

= [ ( 49 - 1 )/7 ]/[ ( 14 + 49 + 1 )/7 ]

= 48/64

After cancellation, we get

= 3/4

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