If k1-tan 27-tan- and k2-sin-cos 3-sin 3-cos 9-sin 9-cos 27- then k1-k2 is-

Consider tan 3θ tanθ=sin3θ/cos 3θ sinθ/cosθ=sin 2θ/cos 3θcosθ=2sinθ/cos 3θ Similarly, tan 9θ tan 3θ=2sin3θ/cos 9θ and tan 27θ tan 9θ =2sin 9θ/cos ...

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Byju's Answer

Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

If k1=tan 2...

Question

If $k_{1} = tan 27 θ - tan θ$ and $k_{2} = \frac{sin θ}{cos 3 θ} + \frac{sin 3 θ}{cos 9 θ} + \frac{sin 9 θ}{cos 27 θ}$ , then $\frac{k_{1}}{k_{2}}$ is:

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Solution

Consider $tan 3 θ - tan θ = \frac{s i n 3 θ}{cos 3 θ} - \frac{sin θ}{cos θ} = \frac{sin 2 θ}{cos 3 θ cos θ} = 2 \frac{sin θ}{cos 3 θ}$
Similarly, $tan 9 θ - tan 3 θ = 2 \frac{s i n 3 θ}{cos 9 θ}$
and $tan 27 θ - tan 9 θ = 2 \frac{sin 9 θ}{cos 27 θ}$
$∴ \frac{sin θ}{cos 3 θ} + \frac{s i n 3 θ}{cos 9 θ} + \frac{sin 9 θ}{cos 27 θ} = \frac{1}{2} (tan 27 θ - tan θ)$
$\Rightarrow k_{2} = \frac{1}{2} k_{1} \Rightarrow k_{1} = 2 k_{2} \Rightarrow \frac{k_{1}}{k_{2}} = 2$

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If k1=tan27θ−tanθ and k2=sinθcos3θ+sin3θcos9θ+sin9θcos27θ, then k1k2 is:

Consider tan3θ−tanθ=sin3θcos3θ−sinθcosθ=sin2θcos3θcosθ=2sinθcos3θSimilarly, tan9θ−tan3θ=2sin3θcos9θand tan27θ−tan9θ=2sin9θcos27θ ∴sinθcos3θ+sin3θcos9θ+sin9θcos27θ=12(tan27θ−tanθ) ⇒k2=12k1⇒k1=2k2⇒k1k2=2

If $k_{1} = tan 27 θ - tan θ$ and $k_{2} = \frac{sin θ}{cos 3 θ} + \frac{sin 3 θ}{cos 9 θ} + \frac{sin 9 θ}{cos 27 θ}$ , then $\frac{k_{1}}{k_{2}}$ is: