^{-1}( θ)= [AP EAPCET 23 May 2025 S2]

Correct Answer is : cosh⁡−1(sec⁡θ)^{-1}( θ)cosh−1(secθ)

Correct Answer is : cosh⁡−1(sec⁡θ)^{-1}( θ)cosh−1(secθ)

Test Index

Inverse Trigonometric Functions

Section: Mathematics

Question : 8 of 33

Marks: +1, -0

\tanh^{-1}(\sin \theta)=

Solution:

\because \tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \;(\text{ for }|x|<1)

\;\;\;\;\; \text{so, } \tanh^{-1}(\sin \theta)=\frac{1}{2}\ln\left(\frac{1+\sin \theta}{1-\sin \theta}\right)

\frac{1+\sin \theta}{1-\sin \theta}=\frac{(1+\sin \theta)^2}{1-\sin^2 \theta}=\left(\frac{1+\sin \theta}{\cos \theta}\right)^2

=\frac{1}{2}\ln\left(\frac{1+\sin \theta}{\cos \theta}\right)^2

=\ln\left(\frac{1+\sin \theta}{\cos \theta}\right) \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

\text{ and } \cosh^{-1}(x)=\ln\left(x+\sqrt{x^2-1}\right)

\therefore \cosh^{-1}(\sec \theta)=\ln\left(\frac{1}{\cos \theta}+\sqrt{\frac{1-\cos^2 \theta}{\cos^2 \theta}}\right)

=\ln\left(\frac{1+\sin \theta}{\cos \theta}\right) \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)

By comparing Eqs. (i) and (ii), we get

\tanh^{-1}(\sin \theta)=\cosh^{-1}(\sec \theta)

Go to Question:

Prev Question

Next Question