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Inverse Trigonometric Functions question

2022 · 25 Jun · Shift 2 · Q35
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  5. /2022 · 25 Jun · Shift 2 · Q35

Inverse Trigonometric Functions question

2022 · 25 Jun · Shift 2 · Q35

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The value of tan⁡−1(cos⁡(15π4)−1sin⁡(π4)){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)tan−1(sin(4π​)cos(415π​)−1​) is equal to :
  1. A
    −π4- {\pi \over 4}−4π​
  2. B
    −π8- {\pi \over 8}−8π​
  3. C
    −5π12- {{5\pi } \over {12}}−125π​
  4. D
    −4π9- {{4\pi } \over 9}−94π​
View written solutionFree

Correct answer: B

  1. Evaluate the trigonometric values inside the inverse tangent.

We need an−1(cos⁡(15π4)−1sin⁡(π4)). an^{-1}\left(\frac{\cos\left(\frac{15\pi}{4}\right)-1}{\sin\left(\frac{\pi}{4}\right)}\right).an−1(sin(4π​)cos(415π​)−1​).

  1. Simplify cos⁡(15π4)\cos\left(\frac{15\pi}{4}\right)cos(415π​).

Using periodicity of cosine, 15π4=2π+7π4,\frac{15\pi}{4}=2\pi+\frac{7\pi}{4},415π​=2π+47π​, so cos⁡(15π4)=cos⁡(7π4)=22.\cos\left(\frac{15\pi}{4}\right)=\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt2}{2}.cos(415π​)=cos(47π​)=22​​.

Also, sin⁡(π4)=22.\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}.sin(4π​)=22​​.

  1. Substitute these values.

The argument of tan⁡−1\tan^{-1}tan−1 becomes 22−122.\frac{\frac{\sqrt2}{2}-1}{\frac{\sqrt2}{2}}.22​​22​​−1​.

Multiply numerator and denominator relation carefully:

=\frac{\sqrt2-2}{\sqrt2} =1-\frac{2}{\sqrt2} =1-\sqrt2.$$ So the expression is $$\tan^{-1}(1-\sqrt2).$$ 4. Recognize the standard tangent value. We know $$\tan\left(\frac{\pi}{8}\right)=\sqrt2-1.$$ Hence, $$1-\sqrt2=-(\sqrt2-1)=-\tan\left(\frac{\pi}{8}\right).$$ Therefore, $$\tan^{-1}(1-\sqrt2)=\tan^{-1}\left(-\tan\frac{\pi}{8}\right).$$ Since the principal value range of $\tan^{-1}x$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $-\frac{\pi}{8}$ lies in this interval, $$\tan^{-1}(1-\sqrt2)=-\frac{\pi}{8}.$$ 5. Compare with the options. Thus the correct option is $$\boxed{\text{B }\left(-\frac{\pi}{8}\right)}.$$
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