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Inverse Trigonometric Functions

Section: Mathematics

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Question : 15 of 33

Marks: +1, -0

If

\operatorname{sech}^{-1} x = \log 2

\operatorname{cosech}^{-1} y = -\log 3

(x+y)=

[AP EAPCET 22 May 2025 S2]

$\; \frac{1}{6}$
$\; \frac{1}{20}$
$6$
$20$

Solution:

\; \operatorname{sech}^{-1} x = \log 2

\; {\; \text{ and } \; \operatorname{cosech}^{-1} y = -\log 3}^{\operatorname{sech}^{-1} x = \log 2}

\; \Rightarrow \;\; \log \left( \frac{1+\sqrt{1-x^2}}{x} \right) = \log 2

\; \Rightarrow \;\; \frac{1+\sqrt{1-x^2}}{x}=2

\; \Rightarrow \;\; 1+\sqrt{1-x^2}=2x

\; \sqrt{1-x^2}=2x-1

Squaring both sides, we get

\; \Rightarrow \;\; 1-x^2=4x^2+1-4x

\; \Rightarrow \;\; 5x^2-4x=0

\; \Rightarrow \;\; x(5x-4)=0

\; \;\; x=\; \frac{4}{5}

\; \because \operatorname{cosech}^{-1} y = -\log 3

\; \Rightarrow \;\; \log \left( \frac{1+\sqrt{1+y^2}}{y} \right) = \log \; \frac{1}{3}

\; \Rightarrow \;\; \frac{1+\sqrt{1+y^2}}{y} = \; \frac{1}{3}

\; \Rightarrow \;\; 1+\sqrt{1+y^2} = \frac{y}{3}

\; \Rightarrow \;\; \sqrt{1+y^2} = \frac{y}{3} - 1

Squaring both sides, we get

\; \Rightarrow \;\; 1+y^2 = \frac{y^2}{9} + 1 - \frac{2}{3} y

\; \Rightarrow \;\; \frac{8y^2}{9} + \frac{2}{3} y = 0

\; \Rightarrow \;\; \frac{2}{3} y \left( \frac{4y}{3} + 1 \right) = 0

\; \Rightarrow \;\; y = \frac{-3}{4}

\; \Rightarrow \;\; x+y = \frac{4}{5} - \frac{3}{4} = \frac{16-15}{20}

\; \Rightarrow \;\; x+y = \frac{1}{20}

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