If 2025^(sin^2 x) − 2025^(cos^2 x) = (√(2025)) then 2025^(cos2x) + (1/(2025^(cos2x) )) = ?


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Question Number 215640 by BaliramKumar last updated on 12/Jan/25

$I f 2025^{{s i n}^{2} x} - 2025^{{c o s}^{2} x} = \sqrt{2025}$ $t h e n 2025^{c o s 2 x} + \frac{1}{2025^{c o s 2 x}} = ?$
	Answered by Frix last updated on 13/Jan/25

	$t = 2025^{\frac{\cos 2 x}{2}} \Leftrightarrow x = \frac{\cos^{- 1} \frac{2 \ln t}{\ln 2025}}{2}$ $\frac{1}{t} - t = 1 \Rightarrow t = - \frac{1}{2} + \frac{\sqrt{5}}{2}$ $\frac{1}{t^{2}} + t^{2} = 3$
	Answered by som(math1967) last updated on 13/Jan/25

	$2025^{\frac{1 - c o x 2 x}{2}} - 2025^{\frac{1 + c o x 2 x}{2}} = \sqrt{2025}$ $\Rightarrow \sqrt{2025} [2025^{- \frac{c o s 2 x}{2}} - 2025^{\frac{c o s 2 x}{2}}] = \sqrt{2025}$ $\Rightarrow {[2025^{- \frac{c o s 2 x}{2}} - 2025^{\frac{c o s 2 x}{2}}]}^{2} = 1^{2}$ $\Rightarrow 2025^{- c o s 2 x} + 2025^{c o x 2 x} - 2 . {(2025)}^{0} = 1$ $∴ 2025^{c o s 2 x} + \frac{1}{2025^{c o s 2 x}} = 3$
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