Question

solve 17, 19 and 23...
...

Answer 1

some part of question cropped so, I use
my possibility here .

17) x = 2sin∅/( 1+ sin∅ + cos∅)

= 2sin∅( 1+sin∅ -cos∅)/(1+sin∅+cos∅)(1+sin∅ - cos∅)

=2sin∅( 1+sin∅ - cos∅)/{(1+sin∅)² -cos²∅}

{ (1+sin∅)² -cos²∅ = 1+sin²∅+2sin∅-cos²∅ = 2sin²∅ +2sin∅ }

=2sin∅( 1 + sin∅-cos∅)/2sin∅( 1+sin∅)

=( 1 + sin∅ - cos∅)/(1 + sin∅) = x


19 , x = b/a


√{(a + b)/( a - b)} + √{(a - b)/( a + b)}

= √{ ( 1+ b/a)/( 1-b/a)} + √{ ( 1-b/a)/( 1+b/a)}

=√{ 1+x)/( 1-x) } + √{ 1-x)/(1+x}

= { √(x +1)² + √(x -1)²}/√(x² -1)

= 2x/(x² - 1)


23 ) sin∅ + cos∅ = m

sin^6∅ + cos^6∅ = { sin²∅)³ + (cos²∅)³

= (sin²∅ + cos²∅)³ -3sin²∅.cos²∅(sin²∅+ cos²∅)

we know , sin²∅ + cos²∅ = 1

= 1 - 3sin²∅.cos²∅ ------------------(1)


now ,
sin∅ + cos∅ = m

take square both sides

(sin²∅ + cos²∅ )+2sin∅.cos∅ = m²

2sin∅.cos∅ = m² - 1

sin∅.cos∅ = ( m²-1)/ 2

take square both sides

sin²∅.cos²∅ = (m²-1)²/4

put this in eqn (1)

sin^6∅ + cos^6∅ = 1-3sin²∅.cos²∅

= 1 - 3(m² -1 )²/4

= { 4 - 3(m² -1)²}/4 = RHS

Answer 2

some part of question cropped so, I use my possibility here.

17) x = 2sinØ/( 1+ sin Ø + cos)

= - 2sin Ø( 1+sin Ø -cos)/(1+sinØ+cos) (1+sin Ø - Cos )

=2sinø( 1+sino - cosØ){(1+sinØ)? . cosØ}

{ (1+sin)? -cos?Ø = 1+sin?Ø+2sin- = cos?Ø = 2sin?Ø +2sinØ }

=2 sin ( 1 + sin -cos/2 sin ( 1+song)

=(1+ sin Ø - cosØ)/(1+ sinØ) =x

19, x= b/a

V[(a + b)/( a - b)} + V{{a - b)/( a + b)}

= V{ ( 1+ b/a)/( 1-b/a)} + V{ ( 1-b/a)/( 1+b/a)}

=V( 1+x)/( 1-x) } + v{ 1-x)/(1+x}

= { V(x +1)2 + v(x - 1)2}/V1x2 - 1)

= 2x/(x? - 1) =

23) sin Ø + cos = m

sin^60 + cos^6Ø = { sin?Ø)3 + (cos²013

= (sin?Ø + cos?Ø)3 -3sin?Ø.cos?Ø(sin?Ø+ cos?Ø)

we know, sin?Ø + cos20 = 1

= 1 - 3sin?Ø.cos20

now

-(1)

now, sin + cosØ = m

take square both sides

(sin? + cos² )+2sin.cos = m2

2sinØ.cosØ = m? 1

sinØ.cos Ø = (m2-1)/ 2

take square both sides

sin?Ø.cos?Ø = (m²-1)²/4

put this in equation (1)

sin^60 + cos^60 = 1-3sin?Ø.cos

=1 - 3(m? -1)?/4

= {4 - 3(m? -1)?y4 = RHS