Let f- -sin tan-1 sin-cos 2- - where -4-4- Then the value of ddtan- f- is

We have nbsp; f(θ )=sin ( tan^ 1 ( sinθ√(cos 2θ) ) ), =sin ( tan^ 1x√(1 2x^2) );x=sinθ nbsp; =x√(1 x^2) nbsp; =sinθcosθ=tanθ nbsp; nbsp; df(θ )d ta ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

Let fθ =sin t...

Question

Let $f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{cos 2 θ}})),$ where $\frac{π}{4} < θ \frac{π}{4}$ \). Then the value of $\frac{d}{d (tan θ)} (f (θ))$ is

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Solution

We have
$f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{cos 2 θ}})),$
$= sin ({tan}^{- 1} \frac{x}{\sqrt{1 - 2 x^{2}}}); x = sin θ$
$= \frac{x}{\sqrt{1 - x^{2}}}$
$= \frac{sin θ}{cos θ} = tan θ$
$\frac{d f (θ)}{d tan θ} = 1$

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Similar questions

Q. Let

f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{cos 2 θ}}))

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Let f(θ)=sin(tan−1(sinθ√cos2θ)), where π4<θπ4\). Then the value of dd(tanθ)(f(θ)) is

We have f(θ)=sin(tan−1(sinθ√cos2θ)), =sin(tan−1x√1−2x2);x=sinθ =x√1−x2 =sinθcosθ=tanθ df(θ)d tanθ=1

Let $f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{cos 2 θ}})),$ where $\frac{π}{4} < θ \frac{π}{4}$ \). Then the value of $\frac{d}{d (tan θ)} (f (θ))$ is

We have
$f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{cos 2 θ}})),$
$= sin ({tan}^{- 1} \frac{x}{\sqrt{1 - 2 x^{2}}}); x = sin θ$
$= \frac{x}{\sqrt{1 - x^{2}}}$
$= \frac{sin θ}{cos θ} = tan θ$
$\frac{d f (θ)}{d tan θ} = 1$