MCQ

Sum

If a is any real number, the number of roots of \[\cot x - \tan x = a\] in the first quadrant is (are).

#### Options

2

0

1

none of these

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#### Solution

1

Given:

\[\cot x - \tan x = a\]

\[ \Rightarrow \frac{1}{\tan x} - \tan x = a\]

\[ \Rightarrow 1 - \tan^2 x = a \tan x\]

\[ \Rightarrow \tan^2 x + a \tan x - 1 = 0\]

It is a quadratic equation.

If tan x = z, , then the equation becomes

\[z^2 + az - 1 = 0\]

\[\Rightarrow z = \frac{- a \pm \sqrt{a^2 + 4}}{2}\]

\[ \Rightarrow \tan x = \frac{- a \pm \sqrt{a^2 + 4}}{2}\]

\[ \Rightarrow x = \tan^{- 1} \left( \frac{- a \pm \sqrt{a^2 + 4}}{2} \right)\]

There are two roots of the given equation, but we need to find the number of roots in the first quadrant.

There is exactly one root of the equation, that is,

\[ \Rightarrow \tan x = \frac{- a \pm \sqrt{a^2 + 4}}{2}\]

\[ \Rightarrow x = \tan^{- 1} \left( \frac{- a \pm \sqrt{a^2 + 4}}{2} \right)\]

There are two roots of the given equation, but we need to find the number of roots in the first quadrant.

There is exactly one root of the equation, that is,

\[x = \tan^{- 1} \left( \frac{- a + \sqrt{a^2 + 4}}{2} \right)\].

Concept: Trigonometric Equations

Is there an error in this question or solution?

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