If tancos-1x - sin cot-11-2 - then x -

Given that tan{ cos^ 1(x) } = sin ( cot^ 11/2 ) Let cot^ 11/2 = ϕ⇒1/2 = cot ϕ ⇒ sin ϕ = 1/√(1+cot^2 ϕ) = 2/√(5) Let cos^ 1 x = θ⇒ sec θ1/x⇒ tan θ = √(sec^2 θ ...

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

Byju's Answer

Standard XI

Mathematics

Properties Derived from Trigonometric Identities

If tancos-1x ...

Question

If $t a n (c o s^{- 1} x) = s i n (c o t^{- 1} \frac{1}{2}),$ then x =

$\pm \frac{5}{3}$

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

\pm \frac{\sqrt{5}}{3}

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

\pm \frac{5}{\sqrt{3}}

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

None of these

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

Open in App

Solution

The correct option is B $\pm \frac{\sqrt{5}}{3}$
Given that $t a n {c o s^{- 1} (x)} = s i n (c o t^{- 1} \frac{1}{2})$
$L e t c o t^{- 1} \frac{1}{2} = ϕ \Rightarrow \frac{1}{2} = c o t ϕ$
$\Rightarrow s i n ϕ = \frac{1}{\sqrt{1 + c o t^{2} ϕ}} = \frac{2}{\sqrt{5}}$
$L e t c o s^{- 1} x = θ \Rightarrow s e c θ \frac{1}{x} \Rightarrow t a n θ = \sqrt{s e c^{2} θ - 1}$
$\Rightarrow t a n θ = \sqrt{\frac{1}{x^{2}} - 1} \Rightarrow t a n θ = \frac{\sqrt{1 - x^{2}}}{x}$
$S o, t a n {c o s^{- 1} (x)} = s i n (c o t^{- 1} \frac{1}{2})$
$\Rightarrow t a n (t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x}) = s i n (s i n^{- 1} \frac{2}{\sqrt{5}}) \Rightarrow \frac{\sqrt{1 - x^{2}}}{x} = \frac{2}{\sqrt{5}} \Rightarrow \sqrt{(1 - x^{2}) 5} = 2 x$
Squaring both sides, we get $x = \pm \frac{\sqrt{5}}{3} .$

Suggest Corrections

Similar questions

If $t a n (c o s^{- 1} x) = s i n (c o t^{- 1} \frac{1}{2}),$ then x =

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

solve for x the following equations :

(a)

{c o t}^{- 1} x + {s i n}^{- 1} \frac{1}{\sqrt (5)} = \frac{π}{4}

(b)

2 {t a n}^{- 1} (c o s x) = t a n {t a n}^{- 1} (2 c o s e c x)

(c)

t a n ({c o s}^{- 1} x) = s i n ({c o t}^{- 1} \frac{1}{2})

Solution of equation $t a n (c o s^{- 1} x) = s i n [c o t^{- 1} \frac{1}{2}]$ is

Q. If

tan (c o s^{- 1} x) = sin ({sec}^{- 1} (\sqrt{5}))

, then

x =

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

Join BYJU'S Learning Program

If tan(cos−1x)=sin(cot−112), then x =

If $t a n (c o s^{- 1} x) = s i n (c o t^{- 1} \frac{1}{2}),$ then x =