Question

Prove it: (1/tan3A-tanA)-(1/cot3A+cotA)=cot2A

Answer 1

Given:

$\boxed{ \boxed{ \tt\dfrac{1}{ \tan3A - \tan A} - \dfrac{1}{ \cot3A - \cot A} }}$

To Prove:

$\boxed{\boxed{\tt \: \cot2A}}$

Solution:

L.H.S :

$= \tt\dfrac{1}{ \tan3A - \tan A} - \dfrac{1}{ \cot3A - \cot A}$

$= \tt\dfrac{1}{ \tan3A - \tan A} - \dfrac{1}{ \dfrac{1}{ \tan3A} - \dfrac{1}{tanA} }$

$= \tt\dfrac{1}{ \tan3A - \tan A} - \dfrac{1}{ \dfrac{\tan A - \tan3 A}{ \tan3A .\tan A}}$

$= \tt\dfrac{1}{ \tan3A - \tan A} - \dfrac{tan3A.tanA}{tanA -tan3A}$

$= \tt\dfrac{1}{ \tan3A - \tan A} + \dfrac{tan3A.tanA}{tan3A -tanA}$

$= \tt\dfrac{1 + tan3A.tanA}{ \tan3A - \tan A}$

$= \tt \dfrac{ \dfrac{1}{tan3A - tanA} }{1 + tan3A.tanA}$

$= \tt \dfrac{1}{tan(3A - A )}$

$= \tt\dfrac{1}{ \tan 2A}$

$= \tt{ \cot2A} = [R.H.S]$

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$\boxed{\begin{array}{} \rm \sin(a + b) = \sin a. \cos b + \cos a. \sin b \\ \rm \sin(a - b) = \sin a. \cos b - \cos a. \sin b \\ \rm \cos(a + b) = \cos a. \cos b - \sin a. \sin b \\ \rm \cos(a - b) = \cos a. \cos b + \sin a. \sin b \end{array}}$

$\boxed{\begin{array}{} \rm \tan(a + b) = \dfrac{ \tan a + \tan b}{1 - \tan a. \tan b } \\ \\\rm \tan(a - b) = \dfrac{ \tan a - \tan b}{1 + \tan a. \tan b } \end{array}}$

• Sin θ = 1/Csc θ
• Csc θ = 1/Sin θ
• Cos θ = 1/Sec θ
• Sec θ = 1/Cos θ
• tan θ = 1/Cot θ
• Cot θ = 1/tan θ

Pythagorean Trigonometric Identities:

• sin² θ + cos² θ = 1
• 1+tan² θ= sec² θ
• cosec²θ = 1 + cot²θ

Ratio Trigonometric Identities :

• tan θ = Sin θ/Cos θ
• Cot θ = Cos θ/Sin θ