Question

tan^2x=1-e^2 prove secx +tan^3xcosecx=(2-e^2)3/2

Answer 1

$\textbf{Given:}$

$tan^2x=1-e^2$

$\textbf{To prove:}$

$secx+tan^3x\;cosecx=(2-e^2)^{\frac{3}{2}}$

$\text{Consider,}$

$tan^2x=1-e^2$

$\text{Add 1 on bothsides, we get}$

$1+tan^2x=2-e^2$

$sec^2x=2-e^2$.......(1)

$\text{Now,}$

$secx+tan^3x\;cosecx$

$=secx+\frac{sin^3x}{cos^3x}(\frac{1}{sinx})$

$=secx+\frac{sin^2x}{cos^3x}$

$=secx+\frac{sin^2x}{cos^2x}(\frac{1}{cosx})$

$=secx+tan^2x\,secx$

$=secx(1+tan^2x)$

$=secx(sec^2x)$

$=sec^3x$

$=(sec^2x)^{\frac{3}{2}}$

$\text{Using (1)}$

$=(2-e^2)^{\frac{3}{2}}$

$\therefore\boxed{\bf\,secx+tan^3x\;cosecx=(2-e^2)^{\frac{3}{2}}}$

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Answer 2

Step-by-step explanation:

the answer is (2-p^2)^3/2.

I have substituted p in the place of e.

Thank you.

Hope it is useful