## Book: RD Sharma - Mathematics (Volume 1)

### Chapter: 4. Inverse Trigonometric Functions

#### Subject: Maths - Class 12th

##### Q. No. 1 of MCQ

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1
##### Choose the correct answerIf , then x2 =

We are given that,

We need to find the value of x2.

Take,

Multiply on both sides by tangent.

Since, we know that tan(tan-1 x) = x.

So,

Or

Now, we need to simplify it in order to find x2. So, rationalize the denominator by multiplying and dividing by .

Note the denominator is in the form: (x + y)(x – y), where

(x + y)(x – y) = x2 – y2

So,

…(i)

Numerator:

Applying the algebraic identity in the numerator, (x – y)2 = x2 + y2 – 2xy.

We can write as,

Again using the identity, (x + y)(x – y) = x2 – y2.

…(ii)

Denominator:

Solving the denominator, we get

…(iii)

Substituting values of Numerator and Denominator from (ii) and (iii) in equation (i),

By cross-multiplication,

x2 tan α = 1 – √(1 – x4)

√(1 – x4) = 1 – x2 tan α

Squaring on both sides,

[√(1 – x4)]2 = [1 – x2 tan α]2

1 – x4 = (1)2 + (x2 tan α)2 – 2x2 tan α [, (x – y)2 = x2 + y2 – 2xy]

1 – x4 = 1 + x4 tan2 α – 2x2 tan α

x4 tan2 α – 2x2 tan α + x4 + 1 – 1 = 0

x4 tan2 α – 2x2 tan α + x4 = 0

Rearranging,

x4 + x4 tan2 α – 2x2 tan α = 0

x4 (1 + tan2 α) – 2x2 tan α = 0

x4 (sec2 α) – 2x2 tan α = 0 [, sec2 x – tan2 x = 1 1 + tan2 x = sec2 x]

Taking x2 common from both terms,

x2 (x2 sec2 α – 2 tan α) = 0

x2 = 0 or (x2 sec2 α – 2 tan α) = 0

But x2 ≠ 0 as according to the question, we need to find some value of x2.

x2 sec2 α – 2 tan α = 0

x2 sec2 α = 2 tan α

In order to find the value of x2, shift sec2 α to Right Hand Side (RHS).

Putting ,

x2 = 2 sin α cos α

Using the trigonometric identity, 2 sin x cos x = sin 2x.

x2 = sin 2α

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