The general solution of $tan x - sin x = 1 - tan x sin x$

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

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

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x = n π + \frac{π}{4}

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x = n π + \frac{π}{6} x = n π + (- 1)^{n} (- \frac{π}{2})

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Solution

The correct option is A $x = n π + \frac{π}{4} x = n π + (- 1)^{n} (- \frac{π}{2})$
$tan x - sin x = 1 - tan x sin x$
$\frac{sin x}{cos x} - sin x = 1 - \frac{sin x}{cos x} sin x$
$sin x - sin x cos x = cos x - {sin}^{2}$
$sin x + {sin}^{2} x = cos x + cos x sin x$
$sin x (1 + sin x) = cos x (1 + sin x)$
$(sin x - cos x) (1 + sin x) = 0$
$sin x - cos x = 0 | sin x = - 1 = sin (\frac{- π}{2})$
$\Rightarrow tan x = 1 | ∴ x = n π + (- 1)^{n} (\frac{- π}{2})$
$\Rightarrow tan x = tan \frac{π}{4}$
$x = n π + \frac{π}{4}$
OR
$tan x + tan x sin x = 1 + sin x$
$tan x (1 + sin x) = 1 + sin x$
$(1 + sin x) (t a n x - 1) = 0$
$sin x = - 1 O r t a n x = 1$
$x = n π + (- 1)^{n} (\frac{- π}{2}) x = n π + \frac{π}{4}$

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Q. If

x + y = \frac{π}{4}

and

tan x + tan y = 1,

then

(n \in Z)

Q. General solutions of

x

for which

2 sin x + 1 = 0

is :

Find the general solutions of $3 t a n^{2} x - 1 = 0$

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satisfying

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and

| 2 sin x | < 1

is
(where

n \in Z

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The general solution of tanx−sinx=1−tanxsinx

The general solution of $tan x - sin x = 1 - tan x sin x$