Let f-sintan -1sin-cos 2 - -4-4- Then the value of d-dtan-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 XII

Mathematics

Differentiation of Inverse Trigonometric Functions

Let fθ=sintan...

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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Similar questions

Q. Assertion :Let

θ ε (\frac{π}{4}, \frac{π}{2})

, then statement 1

(cos θ)^{sin θ} < (cos θ)^{cos θ} < (sin θ)^{cos θ}

Reason: The equation

e^{sin θ} - e^{- sin θ} = 4

has a unique solution.

Q. Let

z = \frac{cos θ + i sin θ}{c o s θ - i sin θ}, \frac{π}{4} < θ < \frac{π}{2}

Then arg

z

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {sin}^{2} θ & s i n^{2} θ c o s^{2} θ & 1 + c o s^{2} θ & c o s^{2} θ 4 sin 4 θ & 4 sin 4 θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

, then

sin 4 θ

equals to.

$Solve the following equations : (i) c o s θ + c p s 2 θ + c o s 3 θ = 0 (i i) c o s θ + c o s 3 θ - c o s 2 θ = 0 (i i i) s i n θ + s i n 5 θ = s i n 3 θ (i v) c o s θ c o s 2 θ c o s 3 θ = \frac{1}{4} (v) c o s θ + s i n θ = c o s 2 θ + s i n 2 θ (v i) s i n θ + s i n 2 θ + s i n 3 θ = 0 (v i i) s i n θ + s i n 2 θ + s i n 3 θ + s i n 4 θ = 0 (v i i i) s i n 3 θ - s i n θ = 4 c o s^{2} θ - 2 (i x) s i n 2 θ - s i n 4 θ + s i n 6 θ = 0$

Q. If

α

is a root of

25 {cos}^{2} θ + 5 cos θ - 12 = 0, \frac{π}{2} < α < π .

Then the value of

sin 2 α

is equal to

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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