Question

11. if sec tita+ tan tita=x, prove that sin tita=x square -1 by x square +1

Answer 1

Answer:

The value of $Sin\theta$ is $\frac{x^{2}-1}{x^{2}+1}$.

Step-by-step explanation:

It is provided that $Sec\theta+Tan\theta=x$

Simplify this equation as follows:

$Sec\theta+Tan\theta=x\\\frac{1}{Cos\theta}+ \frac{Sin\theta}{Cos\theta}=x\\\frac{1+Sin\theta}{Cos\theta}=x$

Solve the RHS (right hand side) of the to be proved equation as follows:

$\frac{x^{2}-1}{x^{2}+1}=\frac{(1+Sin\theta)^{2}-Cos^{2}\theta}{(1+Sin\theta)^{2}+Cos^{2}\theta}\\=\frac{1+Sin^{2}\theta+2Sin\theta-Cos^{2}\theta}{1+Sin^{2}\theta+2Sin\theta+Cos^{2}\theta}\\=\frac{Sin^{2}\theta+Cos^{2}\theta+Sin^{2}\theta+2Sin\theta-Cos^{2}\theta}{Sin^{2}\theta+Cos^{2}\theta+Sin^{2}\theta+2Sin\theta+Cos^{2}\theta}\\=\frac{2Sin^{2}\theta+2Sin\theta}{2+2Sin\theta} \\=\frac{2Sin\theta(Sin\theta+1)}{2(Sin\theta+1)}\\ =Sin\theta$

Hence proved.