Let f- - sintan-1sin-cos 2- -4-4- Then the value of dd tan- f- is

F(θ )= sin(tan^ 1(sinθ√(cos 2θ))) ⇒ f(θ )= sin(tan^ 1(sinθ√(2cos^2 θ 1))) ⇒ nbsp;f(θ )= sin(sin^ 1(sinθ√(sin^2 θ +2 cos^2 θ 1))) ⇒ nbsp;f(θ )= sin(sin^ 1( ...

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Standard XIII

Mathematics

Differentiation of Inverse Trigonometric Functions

Let fθ = sint...

Question

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

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Solution

$f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{cos 2 θ}})) \Rightarrow f (θ) = sin ({tan}^{- 1} (\frac{sin θ}{\sqrt{2 {cos}^{2} θ - 1}})) \Rightarrow f (θ) = sin ({sin}^{- 1} (\frac{sin θ}{\sqrt{{sin}^{2} θ + 2 {cos}^{2} θ - 1}})) \Rightarrow f (θ) = sin ({sin}^{- 1} (\frac{sin θ}{\sqrt{{cos}^{2} θ}})) \Rightarrow f (θ) = sin ({sin}^{- 1} (tan θ)) = tan θ$
So,
$\frac{d}{(d (tan θ))} (f (θ)) = \frac{d}{d (tan θ)} tan θ = 1$

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Let f(θ)=sin(tan−1(sinθ√cos2θ));−π4<θ<π4, Then the value of d(d(tanθ))(f(θ)) is

f(θ)=sin(tan−1(sinθ√cos2θ))⇒f(θ)=sin(tan−1(sinθ√2cos2θ−1))⇒f(θ)=sin(sin−1(sinθ√sin2θ+2cos2θ−1))⇒f(θ)=sin(sin−1(sinθ√cos2θ))⇒f(θ)=sin(sin−1(tanθ))=tanθ So, d(d(tanθ))(f(θ))=dd(tanθ)tanθ=1

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