If x-co 2- y-2- z-11-sin2-cos2- then tan 3-tan- is equal to

The correct option is A (3x 4)y/(4 3y)x Given #160; x=cosec^2θ, y=^2θ, z=11 sin^2θcos^2θ ⇒ sin^2 θ= 1x,cos^2 θ= 1y,1 sin^2 θcos^2θ= 1z tan ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

If x=co 2θ,...

Question

If $x = c o {sec}^{2} θ, y = {sec}^{2} θ, z = \frac{1}{1 - {sin}^{2} θ {cos}^{2} θ}$ then $\frac{tan 3 θ}{t a n θ}$ is equal to

\frac{(3 x - 4) y}{(4 - 3 y) x}

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\frac{(4 - 3 y) x}{(3 x - 4) y}

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\frac{(4 - 3 x) y}{(4 - 3 y) x}

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(y / 3)^{3 / 2}

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Solution

The correct option is A $\frac{(3 x - 4) y}{(4 - 3 y) x}$
Given $x = c o s e c^{2} θ, y = {sec}^{2} θ, z = \frac{1}{1 - {sin}^{2} θ {cos}^{2} θ}$

$\Rightarrow s i n^{2} θ = \frac{1}{x}, {cos}^{2} θ = \frac{1}{y}, 1 - {sin}^{2} θ {cos}^{2} θ = \frac{1}{z}$

$t a n 3 θ = \frac{s i n 3 θ}{c o s 3 θ} = \frac{3 s i n θ - 4 s i n^{3} θ}{4 c o s^{3} θ - 3 c o s θ} = t a n θ (\frac{3 - 4 s i n^{2} θ}{4 c o s^{2} θ - 3})$

$\Rightarrow \frac{t a n 3 θ}{t a n θ} = \frac{3 - \frac{4}{x}}{\frac{4}{y} - 3} = \frac{(3 x - 4) y}{(4 - 3 y) x}$

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If x=cosec2θ,y=sec2θ,z=11−sin2θcos2θ then tan3θtanθ is equal to

The correct option is A (3x−4)y(4−3y)xGiven x=cosec2θ,y=sec2θ,z=11−sin2θcos2θ⇒sin2θ=1x,cos2θ=1y,1−sin2θcos2θ=1ztan3θ=sin3θcos3θ=3sinθ−4sin3θ4cos3θ−3cosθ=tanθ(3−4sin2θ4cos2θ−3)⇒tan3θtanθ=3−4x4y−3=(3x−4)y(4−3y)x

If $x = c o {sec}^{2} θ, y = {sec}^{2} θ, z = \frac{1}{1 - {sin}^{2} θ {cos}^{2} θ}$ then $\frac{tan 3 θ}{t a n θ}$ is equal to