Question

prove that 1+sectheta-tantheta/1+sectheta+tantheta=1-sintheta/costheta​

Answer 1

$\begin{aligned}&\large \textsf{LHS...}\\ \\ \Longrightarrow\ \ &\dfrac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}\\ \\ \Longrightarrow\ \ &\dfrac{(1+\sec\theta-\tan\theta)\cos\theta}{(1+\sec\theta+\tan\theta)\cos\theta}\\ \\ \Longrightarrow\ \ &\dfrac{\cos\theta+\sec\theta\cdot \cos\theta-\tan\theta\cdot \cos\theta}{\cos\theta+\sec\theta\cdot\cos\theta+\tan\theta\cdot\cos\theta}\\ \\ \Longrightarrow\ \ &\dfrac{\cos\theta+1-\sin\theta}{\cos\theta+1+\sin\theta}\end{aligned}$

$\begin{aligned}\\ \\ \Longrightarrow\ \ &\dfrac{1-\sin\theta+\cos\theta}{1+\sin\theta+\cos\theta}\\ \\ \Longrightarrow\ \ &\dfrac{(1-\sin\theta+\cos\theta)(1-\sin\theta)}{(1+\sin\theta+\cos\theta)(1-\sin\theta)}\\ \\ \Longrightarrow\ \ &\dfrac{(1-\sin\theta+\cos\theta)(1-\sin\theta)}{1-\sin^2\theta+\cos\theta-\sin\theta\cdot\cos\theta}\\ \\ \Longrightarrow\ \ &\dfrac{(1-\sin\theta+\cos\theta)(1-\sin\theta)}{\cos^2\theta+\cos\theta-\sin\theta\cdot\cos\theta}\end{aligned}$

$\begin{aligned}\\ \\ \Longrightarrow\ \ &\dfrac{(1-\sin\theta+\cos\theta)(1-\sin\theta)}{\cos\theta(\cos\theta+1-\sin\theta)}\\ \\ \Longrightarrow\ \ &\dfrac{(1-\sin\theta+\cos\theta)(1-\sin\theta)}{\cos\theta(1-\sin\theta+\cos\theta)}\\ \\ \Longrightarrow\ \ &\dfrac{1-\sin\theta}{\cos\theta}\\ \\ \Longrightarrow\ \ &\large \textsf{...RHS}\end{aligned}$

$\Huge \texttt{\underline{\underline{Hence Proved!}}}$