Question

Based on compound angles class 11th
To prove that​

Answer 1

{ I am taking α = A for the proof. }

Proof :

L.H.S.

= cotA + cot(60 + A) - cot(60 - A)

= cotA + {cot(60 + A) - cot(60 - A)}

$\displaystyle\mathrm{=cotA+\left(\frac{cos(60+A)}{sin(60+A)}-\frac{cos(60-A)}{sin(60-A)}\right)}$

$\displaystyle\mathrm{=cotA+2\left(\frac{sin(60-A)cos(60+A)-cos(60-A)sin(60+A)}{2sin(60+A)sin(60-A)}\right)}$

$\displaystyle\mathrm{=cotA+2\left(\frac{sin(60-A-60-A)}{cos(60+A-60+A)-cos(60+A+60-A)}\right)}$

$\displaystyle\mathrm{=cotA+\frac{2sin(-2A)}{cos2A-cos120}}$

$\displaystyle\mathrm{=cotA+\frac{-2sin2A}{cos2A+\frac{1}{2}}}$

$\displaystyle\mathrm{=cotA-\frac{\frac{4cotA}{1+cot^{2}A}}{\frac{cot^{2}A-1}{cot^{2}A+1}+\frac{1}{2}}}$

$\displaystyle\mathrm{=cotA-\frac{8cotA}{2cot^{2}A-2+cot^{2}A+1}}$

$\displaystyle\mathrm{=cotA-\frac{8cotA}{3cot^{2}A-1}}$

$\displaystyle\mathrm{=\frac{3cot^{3}A-cotA-8cotA}{3cot^{2}A-1}}$

$\displaystyle\mathrm{=3\left(\frac{cot^{3}A-3cotA}{3cot^{2}A-1}\right)}$

= 3 cot3A = R.H.S.

Hence, proved.

Useful rules :

• $\displaystyle\mathrm{cotA=\frac{cosA}{sinA}}$

• sin(A-B)=sinA cosB - cosA sinB

• 2 sinA sinB = cos(A - B) - cos(A + B)

• sin(- A) = - sinA

• cos120° = cos(90° + 30°) = - sin30 = - 1/2

• $\displaystyle\mathrm{sin2A=\frac{2cotA}{1+cot^{2}A}}$

• $\displaystyle\mathrm{cos2A=\frac{cot^{2}A-1}{cot^{2}A+1}}$

• $\displaystyle\mathrm{cot3A=\frac{cot^{3}A-3cotA}{3cot^{2}A-1}}$

Answer 2

Answer

= cotA + cot(60 + A) - cot(60 - A)

= cotA + {cot(60 + A) - cot(60 - A)}

$\displaystyle\mathrm{=cotA+\left(\frac{cos(60+A{sin(60+A)}-\frac{cos(60-A)}{sin(60-A)}\right)}=cotA+( sin(60+A)cos(60+A)$

− sin(60−A)

cos(60−A)

[tex\displaystyle\mathrm{=cotA+2\left( frac{sin(60A)cos(60+A)cos(60A)sin(60+A) }{2sin(60+A)sin(60-A)}\right)}=cotA+2( 2sin(60+A)sin(60−A)

sin(60−A)cos(60+A−cos(60−A)sin(60+A) )[/tex]

[tex\displaystyle\mathrm{=cotA+2\left(\frac{sin(60-A-60-A{cos(60+A-60+A)-cos(60+A+60-A)}\right)}=cotA+2( cos(60+A−60+A)−cos(60+A+60−A)sin(60−A−60−A))[/tex]

$\displaystyle\mathrm{=cotA+\frac{2sin(-2A)}{cos2A-cos120}}=cotA+ </p><p>cos2A−cos1202sin(−2A)$

$\displaystyle\mathrm{=cotA+\frac{-2sin2A}{cos2A+\frac{1}{2}}}=cotA+ cos2A+ 21 −2sin2A$

[tex\displaystyle\mathrm{=cotA-\frac{\frac{4cotA}{1+cot^{2}A}}{\frac{cot^{2}A-1}{cot^{2}A+1}+\frac{1}{2}}}=cotA− cot 2

A+1cot 2 A−1 + 211+cot 2 A4cotA[/tex]

[tex\displaystyle\mathrm{=cotA-\frac{8cotA}{2cot^{2}A-2+cot^{2}A+1}}=cotA− 2cot 2A−2+cot 2 A+18cotA[/tex]

[tex\displaystyle\mathrm{=cotA\frac{8cotA}{3cot^{2}A-1}}=cotA− 3cot 2A−1 8cotA[/tex]

[tex\displaystyle\mathrm{=\frac{3cot^{3}A-cotA-8cotA}{3cot^{2}A-1}}= 3cot 2

A−13cot 3 A−cotA−8cotA[/tex]

[tex\displaystyle\mathrm{=3\left(\frac{cot^{3}A-3cotA}{3cot^{2}A-1}\right)}=3(

3cot 2 A−1cot 3A−3cotA)[/tex]

= 3 cot3A = R.H.S.

[tex\displaystyle\mathrm{cotA=\frac{cosA}{sinA}}cotA=sinAcosA[/tex]

sin(A-B)=sinA cosB - cosA sinB 2 sinA sinB = cos(A - B) - cos(A + B) sin(- A) = - sinA

cos120° = cos(90° + 30°) = - sin30 = - 1/2

[tex\displaystyle\mathrm{sin2A=\frac{2cotA}{1+cot^{2}A}}sin2A=1+cot 2A2cotA

[tex\displaystyle\mathrm{cos2A=\frac{cot^{2}A-1}{cot^{2}A+1}}cos2A= cot 2 A+1cot 2 A−1[/tex]

[tex\displaystyle\mathrm{cot3A=\frac{cot^{3}A-3cotA}{3cot^{2}A-1}}cot3A=

3cot 2 A−1cot 3A−3cotA[/tex]