Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Find the number of values of

x

satisfying simultaneously

\sin^{-1}x=2\tan^{-1}x

and

\tan^{-1}\sqrt{x(x-1)}+\csc^{-1}\sqrt{1+x-x^{2}}=\dfrac{\pi}{2}

Solution

Answer: 2

Step 1: For the second equation, rewrite using

\csc^{-1}u=\sin^{-1}\dfrac{1}{u}

or equivalently using the identity

\tan^{-1}a+\csc^{-1}b=\dfrac{\pi}{2}

when

a=\sqrt{b^{2}-1}

. Equivalently:

\tan^{-1}\sqrt{x^{2}-x}+\csc^{-1}\sqrt{1-(x^{2}-x)}=\dfrac{\pi}{2}

. Step 2: For domain:

\sqrt{x^{2}-x}

requires

x^{2}-x\ge 0

, and

\csc^{-1}u

requires

|u|\ge 1

, so

\sqrt{1-(x^{2}-x)}\ge 1

, i.e.,

1-(x^{2}-x)\ge 1

, giving

x^{2}-x\le 0

. Step 3: Combining:

x^{2}-x=0

, so

x=0

x=1

. Step 4: Check both in the first equation

\sin^{-1}x=2\tan^{-1}x

: For

x=0

\sin^{-1}0=0

and

2\tan^{-1}0=0

. Equal. ✓ For

x=1

\sin^{-1}1=\dfrac{\pi}{2}

and

2\tan^{-1}1=2\cdot\dfrac{\pi}{4}=\dfrac{\pi}{2}

. Equal. ✓ Step 5: Both values satisfy the system. Answer:

2

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.