Question

Given that tan theta=1/root 3 , what is the value of cosec square theta-sec square theta/cosec square theta +sec square theta?

Answer 1

Hi ,

Here I used A instead of theta.

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

we know that,

1 ) tan A = 1/Cot A

2 ) cosec² A = 1 + cot² A

3 ) sec² A = 1 + tan² A

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

tan A = 1/√3 ----( 1 )

cot A = √3 ------( 2 )

( cosec² A - sec² A )

= 1 + cot² A - ( 1 + tan² A )

= 1 + cot² A - 1 - tan² A

= cot² A - tan² A

= ( √3 )² - ( 1/√3 )²

= 3 - 1/3

= ( 9 - 1 ) / 3

= 8/3 --- ( 3 )

cosec² A + sec² A

= 1 + cot² A + 1 + tan² A

= 2 + cot² A + tan² A

= 2 + ( √3 )² + ( 1/√3 )²

= 2 + 3 + 1/3

= 5 + 1/3

= ( 15 + 1 )/3

= 16/3 ----( 4 )

Therefore ,

required value

= ( 3 ) / ( 4 )

= ( 8/3 ) / ( 16/3 )

= 8/16

= 1/2

I hope this helps you.

:)

Answer 2

Here I used A instead of theta.

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

we know that,

1 ) tan A = 1/Cot A

2 ) cosec² A = 1 + cot² A

3 ) sec² A = 1 + tan² A

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

tan A = 1/√3 ----( 1 )

cot A = √3 ------( 2 )

( cosec² A - sec² A )

= 1 + cot² A - ( 1 + tan² A )

= 1 + cot² A - 1 - tan² A

= cot² A - tan² A

= ( √3 )² - ( 1/√3 )²

= 3 - 1/3

= ( 9 - 1 ) / 3

= 8/3 --- ( 3 )

cosec² A + sec² A

= 1 + cot² A + 1 + tan² A

= 2 + cot² A + tan² A

= 2 + ( √3 )² + ( 1/√3 )²

= 2 + 3 + 1/3

= 5 + 1/3

= ( 15 + 1 )/3

= 16/3 ----( 4 )

Therefore ,

required value

= ( 3 ) / ( 4 )

= ( 8/3 ) / ( 16/3 )

= 8/16

= 1/2

I hope this helps you.

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