Question

Prove that tan 9° - tan 27° - cot 27° + cot 9° = 4

Answer 1

LHS = tan9° - tan27° - cot27° + cot9°

= (tan9° + cot9°) - (tan27° + cot27°)

= (sin9°/cos9° + cos9°/sin9°) - (sin27°/cos27° + cos27°/sin27°)

= (sin²9° + cos²9°)/sin9°cos9° - (sin²27° + cos²27°)/sin27° cos27°

= 1/sin9°cos9° - 1/sin27°cos27°

= 2/(2sin9°cos9°) - 2/(2sin27°cos27°)

= 2/sin18° - 2/sin54° [ as we know , sin2x = 2sinxcosx]

= 2 [ (sin54° - sin18°)/sin18°.sin54°]

use formula, sinX - sinY = 2cos(X + Y)/2sin(X - Y)/2

= 2[(2cos36°sin18°)/sin18°sin54°]

= 4[ cos36°/cos(90° - 36°) ]

= 4[ cos36°/cos36° ]

= 4 = RHS

Answer 2

HELLO DEAR,



tan9° - tan27° - cot27° + cot9°

=> (tan9° + cot9°) - (tan27° + cot27°)

=> (sin9°/cos9° + cos9°/sin9°) - (sin27°/cos27° + cos27°/sin27°)

=> (sin²9° + cos²9°)/sin9°cos9° - (sin²27° + cos²27°)/sin27° cos27°

=> 1/sin9°cos9° - 1/sin27°cos27°

=> 2/(2sin9°cos9°) - 2/(2sin27°cos27°)

=> 2/sin18° - 2/sin54° [ as , sin2θ = 2sinθcosθ]

=> 2 [ (sin54° - sin18°)/sin18°.sin54°]

use formula, sinA - sinB = 2cos(A + B)/2sin(A - B)/2

=> 2[(2cos36°sin18°)/sin18°sin54°]

=> 4[ cos36°/cos(90° - 36°) ]

=> 4[ cos36°/cos36° ]

=> 4


I HOPE IT'S HELP YOU DEAR,
THANKS