Question

Please help me answer this question

Answer 1

Explanation:

We know rhat

`\sin( (x)/(2) ) = \sqrt{ (1 - \cos(x) )/(2) }`

whenever x lies from 90 to 180.

If

`\csc(x) = 9`

Using Reciprocal Identities

`\sin(x) = (1)/(9)`

Using Pythagorean Identity,

`\sin {}^(2) (x) + \cos {}^(2) (x) = 1`

`( (1)/(9) ) {}^(2) + \cos {}^(2) (x) = 1`

`\cos {}^(2) (x) = (80)/(81)`

`\cos(x) = (4 √(5) )/(9)`

Cosine is negative in when x lies between 90 and 180 so

`\cos(x) = - (4 √(5) )/(9)`

So

`\sin( (x)/(2) ) = \sqrt{ (1 + (4 √(5) )/(9) )/(2) }`

`\sin( (x)/(2) ) = \sqrt{ (9 + 4 √(5) )/(18) }`

For cosine remember

`\cos( (x)/(2) ) = \sqrt{ (1 + \cos(x) )/(2) }`

Cosine is negative in second quadrant so

`\cos( (x)/(2) ) = - \sqrt{ (1 + \cos(x) )/(2) }`

`\cos( (x)/(2) ) = - \sqrt{ (1 - (4 √(5) )/(9) )/(2) }`

`\cos( (x)/(2) ) = - \sqrt{ (9 - 4 √(5) )/(18) }`

For tangent,

`\tan( (x)/(2) ) = ( \sin( (x)/(2) ) )/( \cos( (x)/(2) ) )`

`\frac{ \sqrt{ (1 - \cos(x) )/(2) } }{ \sqrt{ (1 + \cos(x) )/(2) } }`

`= ( √(1 - \cos(x) ) )/( √(1 + \cos(x) ) )`

Tangent is negative over 90<x<180 so

`- ( √(1 - \cos(x) ) )/( √(1 + \cos(x) ) )`

`- \sqrt{ (1 + (4 √(5) )/(9) )/(1 - (4 √(5) )/(9) ) }`

`- \sqrt{ ( (9 + 4 √(5) )/(9) )/( (9 - 4 √(5) )/(9) ) }`

`- \sqrt{ (9 + 4 √(5) )/(9 - 4 √(5) ) }`

`- \sqrt{ \frac{1}{(9 - 4 \sqrt{5) {}^(2) } } }`

`- (1)/(9 - 4 √(5) )`

so

`\tan( (x)/(2) ) = - (1)/( 9 + 4 √(5) )`

or

`\tan( (x)/(2) ) = - 9 + 4 √(5)`