When ( sin 9 θ/ cos 27 θ)+( sin 3 θ/ cos 9 θ)+( sin θ/ cos 3 θ)=k ( tan 27 θ- tan θ) is defined, then k=

Question

When ( sin 9 θ/ cos 27 θ)+( sin 3 θ/ cos 9 θ)+( sin θ/ cos 3 θ)=k ( tan 27 θ- tan θ) is defined, then k= (A) (π/2) (B) -(1/2) (C) 1 (D) (π/4)

Tardigrade Admin · Accepted Answer

Given, ( sin 9 θ/ cos 27 θ)+( sin 3 θ/ cos 9 θ)+( sin θ/ cos 3 θ)=k ( tan 27 θ- tan θ) ∵ LHS =( sin 9 θ cos 9 θ+ sin 3 θ cos 27 θ/ cos 27 θ cos 9 θ)+( sin θ/ cos 3 θ) =( sin 18 θ+ sin 30 θ- sin 24 θ/2 cos 9 θ cos 27 θ)+( sin θ/ cos 3 θ) [ sin 21 θ+ sin 5 θ+ sin 33 θ+ sin 27 θ- sin 27 θ = (- sin 21 θ+4 sin θ cos 9 θ cos 27 θ]/4 cos 3 θ cos 9 θ cos 27 θ) = ( sin 15 θ+ sin 33 θ+4 sin θ cos 9 θ cos 27 θ/4 cos 3 θ cos 9 θ cos 27 θ) = (2 sin 24 θ cos 9 θ+4 sin θ cos 9 θ cos 27 θ/4 cos 3 θ cos 9 θ cos 27 θ) 4 = ( sin 24 θ+2 sin θ cos 27 θ/2 cos 3 θ cos 27 θ) = ( sin (27 θ-3 θ)/2 cos 3 θ cos 27 θ)+( sin θ/ cos 3 θ) =(1/2)[( sin 27 θ/ cos 27 θ)-( sin 3 θ/ cos 3 θ)]+( sin θ/ cos 3 θ) =(1/2) ( sin 27 θ/ cos 27 θ)-(1/2)(( sin 3 θ-2 sin θ/ cos 3 θ)) =(1/2) tan 27 θ-(1/2) (3 sin θ-4 sin 3 θ-2 sin θ/ cos θ(4 cos 2 θ-3)) =(1/2)[ tan 27 θ-( sin θ(1-4 sin 2 θ)/ cos θ(1-4 sin 2 θ))] =(1/2)[ tan 27 θ- tan θ] So, k=(1/2)

Q. When $\frac{s i n 9 θ}{c o s 27 θ} + \frac{s i n 3 θ}{c o s 9 θ} + \frac{s i n θ}{c o s 3 θ} = k$
$(tan 27 θ - tan θ)$ is defined, then $k =$

$\frac{π}{2}$

$- \frac{1}{2}$

1

$\frac{π}{4}$

Q. When cos27θsin9θ​+cos9θsin3θ​+cos3θsinθ​=k (tan27θ−tanθ) is defined, then k=

2π​

−21​

1

4π​

Q. When $\frac{s i n 9 θ}{c o s 27 θ} + \frac{s i n 3 θ}{c o s 9 θ} + \frac{s i n θ}{c o s 3 θ} = k$
$(tan 27 θ - tan θ)$ is defined, then $k =$

$\frac{π}{2}$

$- \frac{1}{2}$

$\frac{π}{4}$