If cos-1 1-x2n-1-x2n -ptan-1xqn- 0- x- - Find the value of p-q

The correct option is B 3Since, 0 lt; x lt; ∞ ; 0 lt; x^n lt; ∞ Let, x^n=tanθ⇒θϵ ( 0, π /2 ) #160; ∴cos^ 1 ( 1 x^2n/1+x^2n )=cos ...

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If cos-1 1-...

Question

If ${cos}^{- 1} (\frac{1 - x^{2 n}}{1 + x^{2 n}}) = p {tan}^{- 1} x^{q n}$ , $0 < x < \infty$ . Find the value of $p + q$

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Solution

The correct option is B 3
Since, $0 < x < \infty$ ; $0 < x^{n} < \infty$
Let, $x^{n} = tan θ \Rightarrow θ ϵ (0, π / 2)$

$∴ {cos}^{- 1} (\frac{1 - x^{2 n}}{1 + x^{2 n}}) = {cos}^{- 1} (\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ})$

$= {cos}^{- 1} (cos 2 θ)$

$= 2 θ = 2 {tan}^{- 1} x^{n}$

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If cos−1(1−x2n1+x2n)=ptan−1xqn, 0<x<∞. Find the value of p+q

The correct option is B 3Since, 0<x<∞; 0<xn<∞Let, xn=tanθ⇒θϵ(0,π/2) ∴cos−1(1−x2n1+x2n)=cos−1(1−tan2θ1+tan2θ) =cos−1(cos2θ) =2θ=2tan−1xn

If ${cos}^{- 1} (\frac{1 - x^{2 n}}{1 + x^{2 n}}) = p {tan}^{- 1} x^{q n}$ , $0 < x < \infty$ . Find the value of $p + q$

The correct option is B 3
Since, $0 < x < \infty$ ; $0 < x^{n} < \infty$
Let, $x^{n} = tan θ \Rightarrow θ ϵ (0, π / 2)$

$∴ {cos}^{- 1} (\frac{1 - x^{2 n}}{1 + x^{2 n}}) = {cos}^{- 1} (\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ})$

$= {cos}^{- 1} (cos 2 θ)$

$= 2 θ = 2 {tan}^{- 1} x^{n}$