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Inverse Trigonometric Functions question

2025 · 23 Jan · Shift 1 · Q31
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Inverse Trigonometric Functions question

2025 · 23 Jan · Shift 1 · Q31

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If π2≤x≤3π4\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}2π​≤x≤43π​, then cos⁡−1(1213cos⁡x+513sin⁡x)\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)cos−1(1312​cosx+135​sinx) is equal to
  1. A
    x+tan⁡−1512x+\tan ^{-1} \frac{5}{12}x+tan−1125​
  2. B
    x−tan⁡−143x-\tan ^{-1} \frac{4}{3}x−tan−134​
  3. C
    x+tan⁡−145x+\tan ^{-1} \frac{4}{5}x+tan−154​
  4. D
    x−tan⁡−1512x-\tan ^{-1} \frac{5}{12}x−tan−1125​
View written solutionFree

Correct answer: D

  1. Rewrite the expression inside cos⁡−1\cos^{-1}cos−1.

We have

1213cos⁡x+513sin⁡x.\frac{12}{13}\cos x+\frac{5}{13}\sin x.1312​cosx+135​sinx.

Notice that

cos⁡α=1213,sin⁡α=513\cos\alpha=\frac{12}{13},\qquad \sin\alpha=\frac{5}{13}cosα=1312​,sinα=135​

for

α=tan⁡−1(512),α∈(0,π2).\alpha=\tan^{-1}\left(\frac{5}{12}\right), \quad \alpha\in\left(0,\frac{\pi}{2}\right).α=tan−1(125​),α∈(0,2π​).

Using

cos⁡(x−α)=cos⁡xcos⁡α+sin⁡xsin⁡α,\cos(x-\alpha)=\cos x\cos\alpha+\sin x\sin\alpha,cos(x−α)=cosxcosα+sinxsinα,

we get

1213cos⁡x+513sin⁡x=cos⁡(x−α),\frac{12}{13}\cos x+\frac{5}{13}\sin x=\cos(x-\alpha),1312​cosx+135​sinx=cos(x−α),

where

α=tan⁡−1(512).\alpha=\tan^{-1}\left(\frac{5}{12}\right).α=tan−1(125​).

So the given expression becomes

cos⁡−1(cos⁡(x−α)).\cos^{-1}\left(\cos(x-\alpha)\right).cos−1(cos(x−α)).
  1. Determine the range of x−αx-\alphax−α.

Given

π2≤x≤3π4.\frac{\pi}{2}\le x\le \frac{3\pi}{4}.2π​≤x≤43π​.

Also,

α=tan⁡−1(512),\alpha=\tan^{-1}\left(\frac{5}{12}\right),α=tan−1(125​),

and since α\alphaα is acute,

0<α<π2.0<\alpha<\frac{\pi}{2}.0<α<2π​.

Hence

π2−α≤x−α≤3π4−α.\frac{\pi}{2}-\alpha \le x-\alpha \le \frac{3\pi}{4}-\alpha.2π​−α≤x−α≤43π​−α.

Now,

π2−α>0\frac{\pi}{2}-\alpha>02π​−α>0

and

3π4−α<π\frac{3\pi}{4}-\alpha<\pi43π​−α<π

because α>0\alpha>0α>0. Thus,

x−α∈[0,π].x-\alpha\in[0,\pi].x−α∈[0,π].
  1. Use the principal value property of cos⁡−1\cos^{-1}cos−1.

For any θ∈[0,π]\theta\in[0,\pi]θ∈[0,π],

cos⁡−1(cos⁡θ)=θ.\cos^{-1}(\cos\theta)=\theta.cos−1(cosθ)=θ.

Since x−α∈[0,π]x-\alpha\in[0,\pi]x−α∈[0,π], we have

cos⁡−1(cos⁡(x−α))=x−α.\cos^{-1}(\cos(x-\alpha))=x-\alpha.cos−1(cos(x−α))=x−α.

Therefore,

cos⁡−1(1213cos⁡x+513sin⁡x)=x−tan⁡−1(512).\cos^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right) = x-\tan^{-1}\left(\frac{5}{12}\right).cos−1(1312​cosx+135​sinx)=x−tan−1(125​).
  1. Match with the options.

This is exactly

x−tan⁡−1(512)\boxed{x-\tan^{-1}\left(\frac{5}{12}\right)}x−tan−1(125​)​

which is Option D.

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