Trigonometry> solve.....

solve..

EDHALA LOKESH , 10 Years ago

Grade 11

3 Answers

PRAPULPODISHETTI

(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).

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=cot4A.

Approved

Last Activity: 10 Years ago

SHANMUKESHWAR

(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.

Last Activity: 10 Years ago

Raghu Vamshi Hemadri

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

Last Activity: 10 Years ago