The value of x -2 - 2 - such that sin x-i cos x-1-i- where i-1- is purely imaginary are given byA- n -4- n - ZB- n -4- n - ZC- n -3- n - ZD- n - n - Z

The correct option is A , nϵZ sin x+i cos x/1+i=(1 i)(sin x+icosx)/(1+i)(1 i) =sin x+cos x+i(cos x sin x)/2 ⇒ sin x+cos x=0 ⇒ tan x= 1 ∴ x = n π π 4 , n ∈ Z ...

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Standard XI

Mathematics

Basic Trigonometric Identities

The value of ...

Question

The value of x $\in (- 2 π, 2 π)$ such that $\frac{s i n x + i c o s x}{1 + i}$ , where $i = \sqrt{- 1}$ , is purely imaginary are given by

, nϵZ

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Solution

The correct option is A
, nϵZ

$\frac{s i n x + i c o s x}{1 + i} = \frac{(1 - i) (s i n x + i c o s x)}{(1 + i) (1 - i)}$
$= \frac{s i n x + c o s x + i (c o s x - s i n x)}{2}$
$\Rightarrow s i n x + c o s x = 0 \Rightarrow t a n x = - 1$
$∴ x = n π - \frac{π}{4}, n \in Z$

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The value of x ∈(−2π,2π) such that sin x+icos x1+i, where i=√−1, is purely imaginary are given by

The correct option is A , nϵZ sin x+i cos x1+i=(1−i)(sinx+icosx)(1+i)(1−i) =sin x+cos x+i(cos x−sin x)2 ⇒sin x+cos x=0⇒tan x=−1 ∴x=nπ−π4, n∈Z

The value of x $\in (- 2 π, 2 π)$ such that $\frac{s i n x + i c o s x}{1 + i}$ , where $i = \sqrt{- 1}$ , is purely imaginary are given by

The correct option is A
, nϵZ

$\frac{s i n x + i c o s x}{1 + i} = \frac{(1 - i) (s i n x + i c o s x)}{(1 + i) (1 - i)}$
$= \frac{s i n x + c o s x + i (c o s x - s i n x)}{2}$
$\Rightarrow s i n x + c o s x = 0 \Rightarrow t a n x = - 1$
$∴ x = n π - \frac{π}{4}, n \in Z$