Question

tan 9° *tan 27° * tan 63° * tan 81° =​

Answer 1

tan9° - tan27° - tan63° + tan81°

= cot81° + tan81° - cot 63° - tan63°

= 1/tan81° + tan81° - 1/tan63° - tan63°

= (1+tan²81°)/tan81° - (1+tan²63°)/tan63°

= sec²81°/tan81° - sec²63°/tan63°

= [(1/cos²81° ) * (cos81°/sin81°)]-[(1/cos²63°) * (cos63°/sin63°)]

= {1/(cos81°sin81°)} - { 1/(cos63°sin63°)}

Since, 2 sinAcosA = sin2A

=> sinAcosA = sin2A/2

So, expression =

{2/sin2*81° } - { 2/ sin2*63° }

= 2/sin162° - 2/sin126°

= 2/sin( pi - 18°) - 2/sin(pi - 54°)

= 2/sin18° - 2/sin54° . . . . . . . . . . . . . . . . .(1)

Now, to get the value of sin18°. . . . .

Let A = 18°

2A + 3A = 90°

2A = 90–3A

=> sin2A = sin(90–3A)

=> sin2A = cos3A

=> 2sinAcosA = 4cos^3 A - 3cosA

2sinA cosA - 4cos^3 A +3cosA = 0

=> cosA ( 2sinA- 4cos²A +3) =0

CosA= 0 ruled out

2SinA -4(1-sin²A) + 3 = 0

=> 4sin²A +2sinA-1=0

=> sinA = (-1-√5)/4) is ruled out

=> sinA = (√5-1)/4

=> sin18° = (√5–1)/4 . . . . . . . . . . . .(2)

Now to get the value of sin54°. . . . . . . . Sin3*18° = 3sin18° - 4sin^3 18°

= 3 * (√5-1)/4 - 4(√5,-1)^3/64

= (√5–1) {48–4(√5–1)²} /64

= (√5–1)(6+2√5) /16

= (4√5+4)/16

=> sin 3*18° = (√5+1)/4 . . . . . . . . . .(3)

Now, plug in these values in exp (1)

We get, 8/(√5–1) - 8/(√5+1)

= ( 8√5 + 8 - 8√5 + 8) / 4

= 16/4

= 4 . . . . . . . . . . . Ans

Answer 2

Hiii...friends,

The answer is here,

Tan9×Tan27×Tan63×Tan81

=> Tan9×Tan (90-63)×Tan63×Tan (90-9)

We know that,

Tan (90-A) = CotA

Then,

=> Tan9×Cot63×Tan63×Cot9

=>( Tan9×Cot9)×(Tan63×Cot63)

Since, CotA×TanA = 1.

=> 1×1

=> 1

:-)Hope it help u.