Question

can someone keep the procedure of thissum?I did this but answer wasn't there in options

Answer 1

Hi ,

x = 5 + 2√6

√x = √ ( 5 + 2√6 )

= √ [ ( √3 )² + (√2 )² + 2 × √3 × √2 ]

= √ ( √3 + √2 )²

= √3 + √2 ----( 1 )

1/√x = 1/( √3 + √2 )

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

= ( √3 - √2 ) / [ ( √3 )² - ( √2 )² ]

= ( √3 - √2 ) / ( 3 - 2 )

= √3 - √2 ----( 2 )

√x + 1/√x = √3 + √2 + √3 - √2

= 2√3 ---( 3 )

Now

Tan theta = 1/2 ( √x + 1/√x )

= 1/2 ( 2√3 )

[ From ( 3 ) ]

Tan theta = √3

= Tan 60°

Therefore ,

Theta = 60°

Now ,

Sec² theta + sin² theta

= Sec² 60° + sin² 60°

= ( 1/2 )² + ( √3 / 2 )²

= 1/4 + 3/4

= ( 1 + 3 ) / 4

= 4/4

= 1

Given options are wrong.

I hope this helps you.

: )