Question 39 of 42advanced🔧 ApplyLong Answer5 marks

Solve the following equation:

2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2\tan^{-1}(\cos x) = \tan^{-1}(2\cosec x)
Correct Answer

x = nπ + π/4, n ∈ Z

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 11 | (Page 14)
For More Understanding

Explanation

This question requires solving an inverse trigonometric equation using the formula 2tan⁡−1(u)=tan⁡−1(2u1−u2)2\tan^{-1}(u) = \tan^{-1}\left(\dfrac{2u}{1-u^2}\right). The key steps involve applying this formula to transform the LHS, equating with RHS, and solving for xx.

Domain constraints must be verified: sin⁡x≠0\sin x \neq 0 (for cosec⁡x\cosec x to be defined) and ∣cos⁡x∣<1|\cos x| < 1 (for the formula to be valid). The solution tan⁡x=1\tan x = 1 gives x=π4+nπx = \dfrac{\pi}{4} + n\pi, which satisfies all constraints.

Solution Steps

  1. Step 1: Apply the formula 2tan⁡−1(u)=tan⁡−1(2u1−u2)2\tan^{-1}(u) = \tan^{-1}\left(\dfrac{2u}{1-u^2}\right) to LHS:

    2tan⁡−1(cos⁡x)=tan⁡−1(2cos⁡x1−cos⁡2x)=tan⁡−1(2cos⁡xsin⁡2x)2\tan^{-1}(\cos x) = \tan^{-1}\left(\dfrac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}\left(\dfrac{2\cos x}{\sin^2 x}\right)

  2. Step 2: Equate both sides:

    tan⁡−1(2cos⁡xsin⁡2x)=tan⁡−1(2cosec⁡x)\tan^{-1}\left(\dfrac{2\cos x}{\sin^2 x}\right) = \tan^{-1}(2\cosec x)

  3. Step 3: Taking tan on both sides:

    2cos⁡xsin⁡2x=2cosec⁡x=2sin⁡x\dfrac{2\cos x}{\sin^2 x} = 2\cosec x = \dfrac{2}{\sin x}

  4. Step 4: Simplify:

    cos⁡xsin⁡2x=1sin⁡x\dfrac{\cos x}{\sin^2 x} = \dfrac{1}{\sin x}

    which gives cos⁡x=sin⁡x\cos x = \sin x (since sin⁡x≠0\sin x \neq 0).

  5. Step 5: Therefore, tan⁡x=1\tan x = 1, giving:

    x=π4+nπ,n∈Zx = \dfrac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}

  6. Step 6: Verify domain: For x=π4+nπx = \dfrac{\pi}{4} + n\pi,

    sin⁡x=±12≠0and∣cos⁡x∣=12<1\sin x = \pm\dfrac{1}{\sqrt{2}} \neq 0 \quad \text{and} \quad |\cos x| = \dfrac{1}{\sqrt{2}} < 1

    ∴ x=nπ+π4,n∈Zx = n\pi + \dfrac{\pi}{4}, \quad n \in \mathbb{Z} is the solution.

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