Question

If sec theta+tan theta=x, obtain the values of sec theta,tan theta, sin theta

Answer 1

$\textbf{Given:}$

$\bf\,x=sec\,\theta+tan\,\theta$........(1)

$\frac{1}{x}=\frac{1}{sec\,\theta+tan\,\theta}{\times}\frac{sec\,\theta-tan\,\theta}{sec\,\theta-tan\,\theta}$

$\frac{1}{x}=\frac{sec\,\theta-tan\,\theta}{sec^2\theta-tan^2\theta}$

$\frac{1}{x}=\frac{sec\,\theta-tan\,\theta}{1}$

$\bf\frac{1}{x}=sec\,\theta-tan\,\theta$.................(2)

$\text{Adding (1) and (2)}$

$x+\frac{1}{x}=sec\,\theta+tan\,\theta+sec\,\theta-tan\,\theta$

$\implies\frac{x^2+1}{x}=2\,sec\,\theta$

$\implies\bf\,sec\,\theta=\frac{x^2+1}{2x}$...........(3)

$\text{Subtracting (2) from (1)}$

$x-\frac{1}{x}=sec\,\theta+tan\,\theta-sec\,\theta+tan\,\theta$

$\implies\frac{x^2-1}{x}=2\,tan\,\theta$

$\implies\bf\,tan\,\theta=\frac{x^2-1}{2x}$...........(4)

$\text{Dividing (4) by (3)}$

$\displaystyle\frac{tan\,\theta}{sec\,\theta}=\frac{\frac{x^2-1}{2x}}{\frac{x^2+1}{2x}}$

$\displaystyle\frac{sin\,\theta/cos\,\theta}{1/cos\,\theta}=\frac{x^2-1}{x^2+1}$

$\implies\displaystyle\bf\,sin\,\theta=\frac{x^2-1}{x^2+1}$

Answer 2

Answer:

I hope it will help you