Prove the following1 sin -12 x -1- x 2-2 tan -1 x - x - - 12 cos -11-x2-1-x2-2 tan -1 x- x - 03 tan -12 x -1- x 2-2 tan -1 x -1- x -1

(1) sin^ 1(2x1+x^2)=2tan^ 1x, |x|≤1 Let tanθ=x⇒θ=tan^ 1(x) Put the value of x in sin^ 1(2x1+x^2) We have nbsp;sin^ 1(2tanθ1+tan^2θ)=sin^ 1(sin2θ) =2θ =2tan^ ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

Prove the fol...

Question

Prove the following
$(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1$
$(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0$
$(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1$

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Solution

$(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1$
Let $tan θ = x \Rightarrow θ = {tan}^{- 1} (x)$
Put the value of $x$ in ${sin}^{- 1} (\frac{2 x}{1 + x^{2}})$
We have
${sin}^{- 1} (\frac{2 tan θ}{1 + {tan}^{2} θ}) = {sin}^{- 1} (sin 2 θ)$
$= 2 θ$
$= 2 {tan}^{- 1} x (Proved)$

$(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0$
Again, put the value of $x$ in ${cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$
We have
${cos}^{- 1} (\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ}) = {cos}^{- 1} (cos 2 θ)$
$= 2 θ$
$= 2 {tan}^{- 1} x (Proved)$

$(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1$
Again, put the value of $x$ in ${tan}^{- 1} (\frac{2 x}{1 - x^{2}})$
We have
${tan}^{- 1} (\frac{2 tan θ}{1 - {tan}^{2} θ}) = {tan}^{- 1} (tan 2 θ)$
$= 2 θ$
$= 2 {tan}^{- 1} x (Proved)$

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Similar questions

Q. If

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

each is equal to

2 {tan}^{- 1} x .

,then show that

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

(b)

s i n [(1 / 5) {c o s}^{- 1} x] = 1

(c)

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

Q. Prove that:

2 {tan}^{- 1} x = {tan}^{- 1} \frac{2 x}{1 - x^{2}}

Q. If

x \geq 1, t h e n 2 T a n^{- 1} x + S i n^{- 1} (\frac{2 x}{1 + x^{2}})

=___

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

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Prove the following (1) sin−1(2x1+x2)=2tan−1x, |x|≤1 (2) cos−1(1−x21+x2)=2tan−1x, x≥0 (3) tan−1(2x1−x2)=2tan−1x, −1<x<1

Prove the following
$(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1$
$(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0$
$(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1$