Math, asked by dsk98094, 8 hours ago

prove that
2sinA/cos3A+2sin3A/cos9A+2sin9A/cos27A=tan27A-tanA

Answers

Answered by bhavikmittal005
0

Answer:

Step-by-step explanation: 2 sin A / cos 3A + 2 sin 3A / cos 9A + 2 sin 9A / cos 27A = tan 27A - tan A

2 sin A/cos 3A

Multiply with cos A to Numerator and denominator

= 2 sin A cos A/cos 3A cos A  

Formula (sin 2A = 2 sin A cos A)

= sin 2A / cos 3A cos A  

Formula [ (sin a-b) / cos a cos b = tan a - tan b ]

So,  

= sin 3A - A / cos 3A cos A

= tan 3A - tan A----> Equation 1

Apply same procedure we get

2 sin3A / cos 9A = tan 9A - tan 3A---> Equation 2

2 sin 9A / cos 27A = tan 27A - tan 9A----> Equation 3

Add Equation 1,2 and 3

tan 3A - tan A + tan 9A - tan 3A + tan 27A - tan 9A = tan 27A - tan A

2 sin A / cos 3A + 2 sin 3A / cos 9A + 2 sin 9A / cos 27A = tan 27A - tan A

Hence it has proven.

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