Question

prove that cosπ/16×cot2π/16×cot3π/16×cot7/16​

Answer 1

an²(π/16) + tan²(7π/16)

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)

= [tan(π/16) + tan(7π/16)]² − 2

$\huge \star \: \: \: similarly -$

tan²(3π/16) + tan²(5π/16) = [tan(3π/16) + tan(5π/16)]² − 2

tan²(2π/16) + tan²(6π/16) = [tan(2π/16) + tan(6π/16)]² − 2

• Now,

[tan(π/16) + tan(7π/16)]²

= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]²

= 1/((sin(7π/16) cos(7π/16))²

= 4/sin²(π/8)

Similarly,

[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)

[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,

tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) tan²(5π/16) + tan²(6π/16) + tan²(7π/16)

• = 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4

= 4/sin²(π/8) + 4/sin²(6π/16) + 3

= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3

= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3

= 16/sin²(π/4) + 3

= 35

an²(π/16) + tan²(7π/16)

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)

= [tan(π/16) + tan(7π/16)]² − 2