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solve..

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Grade:11

3 Answers

PRAPULPODISHETTI
669 Points
5 years ago
(sinA−sin 5A+sin 9A−sin 13A )÷(cos A−cos 5A−cos 9A+cos 13A).
sinA+sin 9A-(sin 5A+sin 13A)÷cos A+cos 13A-(cos 5A+cos 9A).
...
=cot4A.
SHANMUKESHWAR
461 Points
5 years ago
 
(sinA−sin 5A+sin 9A−sin 13A )÷(cos A−cos 5A−cos 9A+cos 13A).
sinA+sin 9A-(sin 5A+sin 13A)÷cos A+cos 13A-(cos 5A+cos 9A)=cot4A.
Raghu Vamshi Hemadri
72 Points
5 years ago
Undoubtedly the value of A is (2n+1) π/4.
SinA-Sin5A+Sin9A-Sin13A/CosA-Cos5A-Cos9A+cos13A=Cot4A
(by applying transformations)
2Cos7A.Sin2A-2Cos7A.Sin6A/2Cos7A.Cos6A-2Cos7A.Cos2A=Cot4A
(by cancelling common factor)
Sin2A-Sin6A/Cos6A-Cos2A=Cot4A
(by applying transformations)
-2Cos4A.Sin2A/2Cos4A.Cos2A=Cot4A
(by cancelling common factor and simplification)
Cot4A.Cot2A= -1
Cos2A=0
2A=(2n+1) π/2
A=(2n+1) π/4

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