Let xi-i-1-2-3- -n be the solutions of the equation tan-1x-1- - x - -2tan-16x- Then 6-i-1nxi is equal to

Case 1: When x gt;0 tan^ 1x+^ 1( x)=2tan^ 1 6x ⇒tan^ 1x+π ^ 1x=2 tan^ 1 6x [∵^ 1 ( x)=π ^ 1x ∀ x ∈ℝ] ⇒π2+2tan^ 1x=2tan^ 1 6x ⇒tan^ 1 6x tan^ 1 x=π4 ⇒tan ^ ...

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Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Let xi,i=1,2,...

Question

Let $x_{i}, i = 1, 2, 3, \dots, n$ be the solutions of the equation ${tan}^{- 1} x + {cot}^{- 1} (- | x |) = 2 {tan}^{- 1} 6 x .$ Then $6 n \sum i = 1 x_{i}$ is equal to

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Solution

Case $1 :$ When $x > 0$
${tan}^{- 1} x + {cot}^{- 1} (- x) = 2 {tan}^{- 1} 6 x$
$\Rightarrow {tan}^{- 1} x + π - {cot}^{- 1} x = 2 {tan}^{- 1} 6 x$
$[∵ {cot}^{- 1} (- x) = π - {cot}^{- 1} x \forall x \in R]$
$\Rightarrow \frac{π}{2} + 2 {tan}^{- 1} x = 2 {tan}^{- 1} 6 x$
$\Rightarrow {tan}^{- 1} 6 x - {tan}^{- 1} x = \frac{π}{4}$
$\Rightarrow {tan}^{- 1} (\frac{6 x - x}{1 + 6 x \cdot x}) = \frac{π}{4}$
$\Rightarrow \frac{5 x}{1 + 6 x^{2}} = 1$
$\Rightarrow 6 x^{2} - 5 x + 1 = 0$
$\Rightarrow x = \frac{1}{2}, \frac{1}{3}$

Case $2 :$ When $x < 0$
${tan}^{- 1} x + {cot}^{- 1} x = 2 {tan}^{- 1} 6 x$
$\Rightarrow {tan}^{- 1} 6 x = \frac{π}{4}$
This is not possible as $x < 0$

$6 n \sum i = 1 x_{i} = 6 (\frac{1}{2} + \frac{1}{3}) = 5$

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Similar questions

The mean deviation for n observations $x_{1}, x_{2}, . . . . x_{n}$ from their mean $¯ ¯¯¯ ¯ X$ is given by

Q. If

X

is the mean of n values x₁, x₂, ......,x_n of variable X, then

\sum_{i = 1}^{n} (x_{i} - X)

is always equal to _____________.

Q. Let

x_{1}, x_{2}, . . ., x_{n}

be n observations and

X

be their arithmetic mean. The standard deviation is given by

(a)

\sum_{i = 1}^{n} {(x_{i} - X)}^{2}

(b)

\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}

(c)

\sqrt{\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}}

(d)

\sqrt{\frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - X^{2}}

Q. Solve the following equations for x:

(i)

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

(ii)

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}

(iii)

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

(iv) 2

\tan^{- 1}

(

\sin

x

)

=

\tan

^{$-$ 1} (2

\sin

x

x \neq \frac{π}{2}

.

(v)

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

(vi)

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

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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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Standard XII Mathematics

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Let xi,i=1,2,3,…,n be the solutions of the equation tan−1x+cot−1(−|x|)=2tan−16x. Then 6n∑i=1xi is equal to

Let $x_{i}, i = 1, 2, 3, \dots, n$ be the solutions of the equation ${tan}^{- 1} x + {cot}^{- 1} (- | x |) = 2 {tan}^{- 1} 6 x .$ Then $6 n \sum i = 1 x_{i}$ is equal to