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Given a relation between sin^-1 x and cos^-1 x, find sin^-1 x and then a trigonometric value

This type of question, asked 3 times: 2021 Paper 20, 2021 Paper 25, 26 Jul 2022. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 20

    64
    If (sin⁡−1x)2−(cos⁡−1x)2=a;0<x<1,a≠0\left( \sin^{- 1}x \right)^{2}-\left( \cos^{- 1}x \right)^{2}=a;0 <x< 1,a\neq 0, then the value of 2x2−12x^{2}- 1 is :

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  2. Question 2

    2021 Paper 25

    70
    If sin⁡−1xa=cos⁡−1xb=tan⁡−1yc;0<x<1\frac{\sin^{- 1}x}{a}=\frac{\cos^{- 1}x}{b}=\frac{\tan^{- 1}y}{c};0 <x< 1, then the value of cos⁡(πca+b)\cos\left( \frac{\pi c}{a+b} \right) is:

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  3. Question 3

    26 Jul 2022

    166
    If 0<x<120 < x <\frac{1}{\sqrt{2}} and sin⁡−1xα=cos⁡−1xβ\frac{\sin^{- 1}x}{\alpha}=\frac{\cos^{- 1}x}{\beta}, then a value of sin⁡(2παα+β)\sin\left( \frac{2\pi\alpha}{\alpha+\beta} \right) is

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