Grade 12th passTrigonometry

When sin9a/cos27a+sin3a/cos9a+sina/cos3a=k(tan27a-tana) is defined,then k =

Krishnaveni

7 Years agoGrade 12th pass

1 Answer

Arun

7 Years ago

(sinA / cos3A)
= (1/2) * [(2 * sinA * cosA) / (cosA * cos3A)]
= (1/2) * [(sin2A) / (cosA * cos3A)]
= (1/2) * [sin(3A - A) / (cosA * cos3A)]
= (1/2) * [(sin3A * cosA - cos3A * sinA) / (cosA * cos3A)]
= (1/2) * [{(sin3A * cosA) / (cosA * cos3A)} - {(cos3A * sinA) / (cosA * cos3A)}]
= (1/2) * [(sin3A/cos3A) - (sinA/cosA)]
= (1/2) * (tan3A - tanA)

so, (sinA / cos3A) = (1/2) * (tan3A - tanA) ---------(1)

Now putting 3A in place of A in equation (1) we get
(sin3A / cos9A) = (1/2) * (tan9A - tan3A) ------------(2)

Again putting 9A in place of A in equation (1) we get
(sin9A / cos27A) = (1/2) * (tan27A - tan9A) ------------(3)

Now doing (1)+(2)+(3) we get
(sinA / cos3A) + (sin3A / cos9A) + (sin9A / cos27A)
= (1/2) * (tan3A - tanA + tan9A - tan3A + tan27A - tan9A)
= (1/2) * (tan27A - tanA)

Regards

Arun