SATHEE: Inverse Circular Functions 2 Question 13

Inverse Circular Functions 2 Question 13

13. Prove that $\cos \tan^{- 1} [\sin (\cot^{- 1} x)] = \sqrt{\frac{x^{2} + 1}{x^{2} + 2}}$ .

(2002, 5M)

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Answer:

Correct Answer: 13. 3

Solution:

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

$\begin{aligned} = \cos \tan^{- 1} \sin \sin^{- 1} \frac{1}{\sqrt{1 + x^{2}}} \\ = \cos \tan^{- 1} \frac{1}{\sqrt{1 + x^{2}}} = \sqrt{\frac{x^{2} + 1}{x^{2} + 2}} = RHS \end{aligned}$

Inverse Circular Functions 2 Question 13

13. Prove that $\cos \tan^{- 1} [\sin (\cot^{- 1} x)] = \sqrt{\frac{x^{2} + 1}{x^{2} + 2}}$ .

Answer:

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Inverse Circular Functions 2 Question 13

13. Prove that cos⁡tan−1⁡[sin⁡(cot−1⁡x)]=x2+1x2+2.

Answer:

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13. Prove that $\cos \tan^{- 1} [\sin (\cot^{- 1} x)] = \sqrt{\frac{x^{2} + 1}{x^{2} + 2}}$ .