If x=cos^-1((1-t^2)/(1+t^2)) ; y=sec^-1((1+t^2)/(1-t^2)), where 0<t<1, then (d y)/(d x) is (a) 1

Name: Problem 129, Chapter 2: IIT JEE Super Course in Mathematics: Calculus
Uploaded: 2022-02-27T20:39:36-08:00
Duration: 1 min 34 s
Channel: Malika Singh
Description: If x=cos^-1((1-t^2)/(1+t^2)) ; y=sec^-1((1+t^2)/(1-t^2)), where 0<t<1, then (d y)/(d x) is (a) 1 (b) 0 (c) -1 (d) (1)/(2)

step 1 The value of x is equivalent to cost inverse 1 minus t square over 1 plus t square and the value

Question

If $x=\cos ^{-1}\left(\frac{1-t^{2}}{1+t^{2}}\right) ; y=\sec ^{-1}\left(\frac{1+t^{2}}{1-t^{2}}\right)$, where $0<t<1$, then $\frac{d y}{d x}$ is (a) 1 (b) 0 (c) $-1$ (d) $\frac{1}{2}$

   If $x=\cos ^{-1}\left(\frac{1-t^{2}}{1+t^{2}}\right) ; y=\sec ^{-1}\left(\frac{1+t^{2}}{1-t^{2}}\right)$, where $0<t<1$, then $\frac{d y}{d x}$ is
(a) 1
(b) 0
(c) $-1$
(d) $\frac{1}{2}$

IIT JEE Super Course in Mathematics: Calculus

Trishna Knowledge… 1st Edition

Chapter 2, Problem 129

Instant Answer

Solved by Expert Malika Singh

02/27/2022

Step 1

Step 1: Given the parametric equations $x=\cos ^{-1}\left(\frac{1-t^{2}}{1+t^{2}}\right)$ and $y=\sec ^{-1}\left(\frac{1+t^{2}}{1-t^{2}}\right)$, we can differentiate both equations with respect to $t$. Show more…

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If $x=\cos ^{-1}\left(\frac{1-t^{2}}{1+t^{2}}\right) ; y=\sec ^{-1}\left(\frac{1+t^{2}}{1-t^{2}}\right)$, where $0<t<1$, then $\frac{d y}{d x}$ is (a) 1 (b) 0 (c) $-1$ (d) $\frac{1}{2}$

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Key Concepts

Inverse Trigonometric Functions

These functions, such as the cosine inverse and secant inverse, relate an angle to its corresponding trigonometric ratio. Their properties and defined ranges are essential for converting between algebraic expressions and angles, and for understanding the behavior and composition of functions.

Trigonometric Identities

Identities like the cosine double angle formula provide a way to express complex trigonometric expressions in simpler forms. Recognizing these patterns helps in rewriting expressions and is key in simplifying the functions for differentiation or solving equations.

Chain Rule in Differentiation

The chain rule is a fundamental technique used to differentiate composite functions. When dealing with inverse trigonometric functions composed with other functions, applying the chain rule correctly allows one to compute the derivatives systematically and accurately.

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