Question

Q.4
If tan 90 = (Where 0 << 6.) Find the value of
18
(3 cosec 30 – 4 sec 30)​

Answer 1

Let 3x=y

Given: tan(9x)=tan(3y)=34

⟹sin(3y)cos(3y)=34

⟹4sin(3y)−3cos(3y)=0

⟹5∗(45∗sin(3y)−35∗cos(3y))=0−−(1)

Let cos3α=45−−(2)

⟹sin3α=35−−(3)

(1)⟹5∗(cos3αsin3y−sin3αcos3y)=0

⟹5∗sin(3α−3y)=0−−(4)

Need:

Let p=3csc3x−4sec3x=3sin3x−4cos3x

⟹p=3siny−4cosy=3cosy−4sinysinycosy

⟹p=5∗(35.cosy−45.siny)sinycosy−−(5)

Substituting (2) and (3) in (5)

⟹p=5∗2∗sin3αcosy−cos3αsiny2sinycosy

⟹p=5∗2∗sin3αcosy−cos3αsiny2sinycosy

⟹p=10∗sin(3α−y)sin2y−−(6)

Following are possibilities for (2):

3(α−y)=0

⟹y=α−−(7a)

Substituting (7a) in (6)

p=10∗sin(3α−α)sin(2α)

⟹p=10∗sin2αsin2α=10−−(8a)

3(α−y)=π

⟹y=α−π3−−(7b)

Substituting (7b) in (6)

p=10∗sin(3α−α+π3)sin2(α−π3)

⟹p=10∗sin(2α+π3)sin(2α−2π3)

⟹p=10∗sin(2α−2π3+π)sin(2α−2π3)

⟹p=10∗(−1)∗sin(2α−2π3)sin(2α−2π3)

⟹p=−10—(8b)

3(α−y)=3π

⟹y=α−π−−(7c)

Substituting (7c) in (6)

p=10∗sin(3α−α+π)sin2(α−π)

⟹p=10∗sin(2α+π)sin(2α−2π)

⟹p=10∗(−1)∗sin(2α)(−1)∗sin(2α)

⟹p=10—(8c)

Ans:

From (8a) , (8b) , (8c) , possible answers are ±10