SATHEE: Inverse Circular Functions 2 Question 6

Inverse Circular Functions 2 Question 6

6. The value of $x$ for which $\sin [\cot^{- 1} (1 + x)] = \cos (\tan^{- 1} x)$ is

(a) $\frac{1}{2}$

(b) 1

(d) $- \frac{1}{2}$

(2004, 1M)

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Answer:

Correct Answer: 6. (d)

Solution:

Given, $\sin [\cot^{- 1} (1 + x)] = \cos (\tan^{- 1} x)$

and we know that,

$\cot^{- 1} θ = \sin^{- 1} \frac{1}{\sqrt{1 + θ^{2}}}$ and $\tan^{- 1} θ = \cos^{- 1} \frac{1}{\sqrt{1 + θ^{2}}}$

From Eq. (i),

$\begin{aligned} \sin \sin^{- 1} \frac{1}{\sqrt{1 + (1 + x)^{2}}} & = \cos \cos^{- 1} \frac{1}{\sqrt{1 + x^{2}}} \\ \Rightarrow & \frac{1}{\sqrt{1 + (1 + x)^{2}}} & = \frac{1}{\sqrt{1 + x^{2}}} \\ \Rightarrow & 1 + x^{2} + 2 x + 1 & = x^{2} + 1 \\ \Rightarrow & x & = - \frac{1}{2} \end{aligned}$

Inverse Circular Functions 2 Question 6

6. The value of $x$ for which $\sin [\cot^{- 1} (1 + x)] = \cos (\tan^{- 1} x)$ is

Answer:

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Inverse Circular Functions 2 Question 6

6. The value of x for which sin⁡[cot−1⁡(1+x)]=cos⁡(tan−1⁡x) is

Answer:

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6. The value of $x$ for which $\sin [\cot^{- 1} (1 + x)] = \cos (\tan^{- 1} x)$ is