Question

Please solve these two questions.

Points =10

Answer 1

1) 3[ sin⁴(3π/2 -x ) + sin⁴( 3π + x) ]-2 [ sin^6(π/2 + x) + sin^6(5π - x ]

first of all you have to know,
sin( 3π/2 + x ) = - cosx
sin( 3π + x) = -sinx
sin(π/2+ x) = cosx
sin( 5π - x ) = sin(4π+π-x ) = sin(π-x) = sinx
now,
3[ ( -cosx)⁴ + ( -sinx)⁴] -2[ (cosx)^6 + ( sinx)^6 ]

= 3[ cos⁴x + sin⁴x ] -2[ sin^6x + cos^6x ]

now ,
use a² + b² = (a + b)²-2ab
and a³ + b³ = (a + b)³ -3ab(a + b)

= 3[{ sin²x + cos²x }² -2sin²x.cos²x ] -2[ { sin²x + cos²x}³ -3sin²x.cos²x(sin²x+cos²x)]
use sin²x + cos²x = 1

= 3[ 1- 2sin²xcos²x] -2[ 1-3sin²x.cos²x]

=3 -6sin²x .cos²x -2 + 6sin²x.cos²x

= 1
here you see 1 is independent of x
hence proved


2) sin⁴∅/a + cos⁴∅/b = 1/( a + b)
b(a + b)sin⁴∅ + a( a + b)cos⁴∅ = ab

{ ab + b²} sin⁴∅ + { a² +ab }cos⁴∅ = ab
divide with cos⁴∅ both sides
( ab + b²) tan⁴∅ + ( ab+ a²) = ab.sec⁴∅
we know,
sec²∅ = 1 + tan²∅
so,
(ab + b²)tan⁴∅ + ( ab + a²) = ab( 1+tan²∅)²

( ab + b²)tan⁴∅ + ( ab + a²) = ab( 1+2tan⁴∅ + tan⁴∅)

( ab + b²)tan⁴∅ + ( ab + a²) = ab +2abtan²∅ + abtan⁴∅

b²tan⁴∅ + ( ab + a²) = ab + 2abtan²∅

b²tan⁴∅ -2abtan²∅ + a² =0

this is just like ( a -b)² = a² + 2ab + b²

so, ( btan²∅ -a)² =0
tan²∅ = a/b
tan∅ = ±√a/√b
let tan∅ = √a/√b

so, sin∅ = √a/√(a + b)
cos∅ = √b/√(a + b)

LHS = (sin∅)^8/a³ + (cos∅)^8/b³
=(√a/√(a + b)}^8/a³ + {√b/√(a + b)}^8/b³
=a⁴/a³(a + b)⁴ + b⁴/b³( a + b)⁴
= a/( a + b)⁴ + b/( a + b)⁴
=(a + b)/( a + b)⁴
=1/(a + b)³ = RHS