l i m x → 1 2 s i n ( c o s − 1 x ) − x 1 − t a n ( c o s − 1 x ) is equal to:

l i m x → 1 2 s i n ( c o s − 1 x ) − x 1 − t a n ( c o s − 1 x ) Let c o s − 1 x = π 4 + θ = l i m θ → ∞ 2 s i n θ − 2 t a n θ ( 1 − t a n θ ) = − 1 2

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$ is equal to:

Option 1 - <math> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>

Option 2 - <math> <mrow> <mo>−</mo> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math>

2 Views|Posted 4 months ago

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Answered by

Raj Pandey | Contributor-Level 9

4 months ago

Correct Option - 4

Detailed Solution:

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$

Let $c o s^{- 1} x = \frac{π}{4} + θ$

= $\underset{θ \to \infty}{l i m} \frac{\sqrt[]{2} s i n θ}{- 2 t a n θ} (1 - t a n θ) = - \frac{1}{\sqrt[]{2}}$

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