Question

If $\tan \theta =\frac{20}{21}$, then prove that $\frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }=\frac{3}{7}$

Answer 1

$\tan \theta =\frac{20}{21}$

As $1+{\tan }^2\theta ={\sec }^2\theta$

$1+\frac{20\times 20}{21\times 21}={\sec }^2\theta$

$1+\frac{400}{441}={\sec }^2\theta$

$\frac{841}{441}={\sec }^2\theta$

$\sec \theta =\frac{29}{21}$

$\cos \theta =\frac{21}{29}...............\left(1\right)$

As $1={\cos }^2\theta +{\sin }^2\theta$

$1=\frac{21\times 21}{29\times 29}+{\sin }^2\theta$

$1=\frac{441}{841}+{\sin }^2\theta$

$1-\frac{441}{841}={\sin }^2\theta$

$\frac{400}{841}={\sin }^2\theta$

$\frac{20}{29}=\sin \theta ................\left(2\right)$

$\frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }............\left(3\right)$

Substituting equation $\left(1\right)$ and $\left(2\right)$ in $\left(3\right)$,

$\frac{1-\frac{20}{29}+\frac{21}{29}}{1+\frac{20}{29}+\frac{21}{29}}=\frac{1+\frac{1}{29}}{1+\frac{41}{29}}=\frac{\frac{30}{29}}{\frac{70}{29}}=\frac{30}{70}=\frac{3}{7}$

$LHS=RHS$

Hence proved.