If-2025-sin-2-x-2025-cos-2-x-2025-then-2025-cos2x-1-2025-cos2x-

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