When sin9θcos27θ+sin3θcos9θ+sinθcos3θ=k(tan27θ-tanθ) is defined, then k=

=sin9xcos27x+sin3xcos9x+sinxcos3x=122sin9xcos9xcos27xcos9x+2sin3xcos3xcos9xcos3x+2sinxcosxcos3xcosx=12sin18xcos27xcos9x+sin6xcos9xcos3x+sin2xcos3xcosx=12sin(27x-9x)cos27xcos9x+sin(9x-3x)cos9xcos3x+sin(3x-x)cos3xcosx=12sin27xcos27x-sin9xcos9x+sin9xcos9x-sin3xcos3x+sin3xcos3x-sinxcosx=12[(tan27x-tan9x)+(tan9x-tan3x)+(tan3x-tanx)]=12[tan27x-tanx]

When sin9θcos27θ+sin3θcos9θ+sinθcos3θ=k(tan27θ-tanθ) is defined, then k=

=sin9xcos27x+sin3xcos9x+sinxcos3x=122sin9xcos9xcos27xcos9x+2sin3xcos3xcos9xcos3x+2sinxcosxcos3xcosx=12sin18xcos27xcos9x+sin6xcos9xcos3x+sin2xcos3xcosx=12sin(27x-9x)cos27xcos9x+sin(9x-3x)cos9xcos3x+sin(3x-x)cos3xcosx=12sin27xcos27x-sin9xcos9x+sin9xcos9x-sin3xcos3x+sin3xcos3x-sinxcosx=12[(tan27x-tan9x)+(tan9x-tan3x)+(tan3x-tanx)]=12[tan27x-tanx]

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

When $\frac{\sin 9 θ}{\cos 27 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin θ}{\cos 3 θ} = k (\tan 27 θ - \tan θ)$ is defined, then k=

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{1}{2}$

$- \frac{1}{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$= \frac{\sin 9 x}{\cos 27 x} + \frac{\sin 3 x}{\cos 9 x} + \frac{\sin x}{\cos 3 x}$

$= \frac{1}{2} [\frac{2 \sin 9 x \cos 9 x}{\cos 27 x \cos 9 x} + \frac{2 \sin 3 x \cos 3 x}{\cos 9 x \cos 3 x} + \frac{2 \sin x \cos x}{\cos 3 x \cos x}]$

$= \frac{1}{2} [\frac{\sin 18 x}{\cos 27 x \cos 9 x} + \frac{\sin 6 x}{\cos 9 x \cos 3 x} + \frac{\sin 2 x}{\cos 3 x \cos x}]$

$= \frac{1}{2} [\frac{\sin (27 x - 9 x)}{\cos 27 x \cos 9 x} + \frac{\sin (9 x - 3 x)}{\cos 9 x \cos 3 x} + \frac{\sin (3 x - x)}{\cos 3 x \cos x}]$

$= \frac{1}{2} [(\frac{\sin 27 x}{\cos 27 x} - \frac{\sin 9 x}{\cos 9 x}) + (\frac{\sin 9 x}{\cos 9 x} - \frac{\sin 3 x}{\cos 3 x}) + (\frac{\sin 3 x}{\cos 3 x} - \frac{\sin x}{\cos x})]$