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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Standard Values

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 $p + q$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

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

$= {cos}^{- 1} (cos 2 θ)$

$= 2 θ = 2 {tan}^{- 1} x^{n}$

Suggest Corrections

Similar questions

Number of greatest binomial coefficients in the expansion of $(1 + x)^{2 n}$ and $(1 + x)^{2 n + 1}$ are p and q respectively (where n is a positive integer). Find the value of p+2q

___

Q. Let the coefficients of

x^{n}

{(1 + x)}^{2 n} & {(1 + x)}^{2 n - 1}

be P and Q, respectively, then

{(\frac{P + Q}{Q})}^{5} =

Q. Let the co-efficient of

x^{n} i n (1 + x)^{2 n} a n d (1 + x)^{2 n - 1}

be P& Q respectively, then

{(\frac{P + Q}{Q})}^{5} =

Q. If

x_{1}, x_{2}, x_{3}, x_{4} . . . . x_{2 n + 1}

are in Arithmetic Progression, then find the value of

[\frac{(x_{2 n + 1} - x_{1})}{(x_{2 n + 1} + x_{1})}] + [\frac{(x_{2 n} - x_{2})}{(x_{2 n} + x_{2})}] . . . . . . . . . . . . . . . . . . . . [\frac{(x_{n + 2 - x_{n}})}{(x_{n + 2} + x_{n})}]

Q. If

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . . . . . + a_{2 n} x^{2 n}

then the value of

a_{0} + a_{3} + a_{6} + . . . . .

is equal to

Join BYJU'S Learning Program

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, $x^{n} = tan θ \Rightarrow θ ϵ (0, π / 2)$

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

$= {cos}^{- 1} (cos 2 θ)$

$= 2 θ = 2 {tan}^{- 1} x^{n}$