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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 25 of 160

Marks: +1, -0

\sin \theta \cosh \alpha = \tan x, \cos \theta \sinh \alpha = \sec x

\cos 2 \theta \cosh 2 \alpha =

Solution:

We have,

\;\; \sin \theta \cosh \alpha = \tan x

\cos \theta \sinh \alpha = \sec x

\cos^2 \theta \sin^2 h \alpha - \sin^2 \theta \cos^2 h \alpha = \sec^2 x - \tan^2 x

\cos^2 \theta \sin^2 h \alpha - \sin^2 \theta \cos^2 h \alpha = 1

\cos^2 \theta \sin^2 h \alpha - \cos^2 h \alpha + \cos^2 \theta \cos^2 h \alpha = 1

\cos^2 \theta (\sin^2 h \alpha + \cos^2 h \alpha) - \cos^2 h \alpha = 1

\cos^2 \theta \cosh 2 \alpha - \cos^2 h \alpha = 1

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

2 \cos^2 \theta \cosh 2 \alpha - \cosh 2 \alpha - 1 = 2

\cosh 2 \alpha (2 \cos^2 \theta - 1) = 3 \Rightarrow \cos 2 \theta \cosh 2 \alpha = 3

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