The number of solutions of the equation tan xsin x - 1 = tan x - sin x, x ∈ [0,2π]is equal to.

Correct answer: B. 2 tan xsin x - tan x + (sin x - 1) = 0 ⇒ (tan x + 1)(sin x - 1) = 0 ⇒ tan x = - 1 or sin x = 1 (Not possible as x ≠ (π)/(2) ) Hence, for~ x = - 1 ~there are& 2 solutions in~[0, 2π]

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The number of solutions of the equation tan xsin x - 1 = tan x -...

The number of solutions of the equation

tan xsin x - 1 = tan x - sin x, \forall x \in \left[0,2\pi\right]is equal to.

Options

A $1$
B $2$
C $3$
D $4$

Correct answer

B. $2$

Step-by-step solution

tan xsin x - tan x + \left(sin x - 1\right) = 0

<br />

⇒ \left(tan x + 1\right)\left(sin x - 1\right) = 0

<br />

⇒ tan x = - 1 or sin x = 1

(Not possible as

x \ne \dfrac{π}{2}

)<br />Hence, for~

\tan x = - 1

~there are&

solutions in~\left[0, 2\pi\right]

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