Question

Prove that: (1/tan3A + tanA) - (1/cot3A + cotA) = cot4A

Answer 1

Question :

Prove that :

$\dfrac{1}{ \tan(3a) + \tan(a) } - \dfrac{1}{ \cot(3a) + \cot(a) } = \cot(4a)$

Trignometric Formula's :

1) tan(A+B)= tanA+tanB/(1-tanAtanB )

2) sin2A = 2 sinA cosA

3)cos2A = cos²A - sin²A

4)tan2A = 2 tanA / (1 - tan²A)

Solution :

LHS

$= \dfrac{1}{ \tan(3a) + \tan(a) } - \dfrac{1}{ \cot(3a) + \cot(a) }$

$= \dfrac{1}{ \tan(3a) + \tan(a) } - \dfrac{1}{ \frac{1}{ \tan(3a) } + \frac{1}{ \tan(a) } }$

$= \dfrac{1}{ \tan(3a) + \tan(a) } - \dfrac{1}{ \frac{ \tan(a) + \tan(3a) }{ \tan(3a) \tan(a) } }$

$= \dfrac{1}{ \tan(3a) + \tan(a) } - \dfrac{ \tan(a) \tan(3a) }{ \tan(3a) + \tan(a) }$

$= \dfrac{1 - \tan(3a) \tan(a) }{ \tan(3a) + \tan(a) }$

$= \dfrac{1}{ \frac{ \tan(3a) + \tan(a) }{1 - \tan(a) \tan(3a) } }$

Now use Formula tan(A+B)= tanA+tanB/(1-tanAtanB )

$= \dfrac{1}{ \tan( 3a + a) }$

$= \dfrac{1}{ \tan(4a) }$

$= \cot(4a)$

RHS = cot 4A

⇒LHS = RHS

Hence proved!

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More Trigonometric Formula's

1. sin²A + cos²A = 1
2. sec²A - tan²A = 1
3. cosec²A - cot²A = 1