Math, asked by HARSHITSRIVASTAVA, 1 year ago

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

Answers

Answered by MaheswariS
14

\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}}}

Find more:

If sinθ+cosθ=x prove that sin6θ+cos6θ=4-3(x²-1)²/4

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Answered by adarsh6279
0

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

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