Question

Prove that
(sinA-cosA+1)/(sinA+cosA-1) = 1/(secA-tanA) using the identity Sec²A = 1+tan²A.
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Answer 1

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Answer 2

$\displaystyle \begin{aligned}&\large \textsf{LHS...}\\ \\ \Longrightarrow\ \ &\frac{\sin A-\cos A+1}{\sin A+\cos A-1}\end{aligned}$

$\text{First we have to multiply both the numerator and denominator with \sec A }$

$\displaystyle \begin{aligned}\Longrightarrow\ \ &\frac{\sin A-\cos A+1}{\sin A+\cos A-1}\\ \\ \Longrightarrow\ \ &\frac{(\sin A-\cos A +1)\sec A}{(\sin A+\cos A-1)\sec A}\\ \\ \Longrightarrow\ \ &\frac{\sin A \cdot \sec A-\cos A \cdot \sec A+\sec A}{\sin A \cdot \sec A+\cos A \cdot \sec A-\sec A}\\ \\ \Longrightarrow\ \ &\frac{\tan A-1+\sec A}{\tan A+1-\sec A}\end{aligned}$

$\text{Now, multiplying both the numerator and the denominator with}\\ \text{ (\sec A-\tan A) ,}$

$\displaystyle \begin{aligned}\Longrightarrow\ \ &\frac{\tan A-1+\sec A}{\tan A+1-\sec A}\\ \\ \Longrightarrow\ \ &\frac{(\tan A+\sec A-1)(\sec A-\tan A)}{(\tan A-\sec A+1)(\sec A-\tan A)}\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\frac{\sec^2A-\tan^2A-\sec A+\tan A}{(\tan A-\sec A+1)(\sec A-\tan A)}\\ \\ \Longrightarrow\ \ &\frac{1+\tan^2A-\tan^2A-\sec A+\tan A}{(\tan A-\sec A+1)(\sec A-\tan A)}\ \ \ \ \ \ \ \ [\because\ \sec^2A=1+\tan^2A]\\ \\ \Longrightarrow\ \ &\frac{\tan A-\sec A+1}{(\tan A-\sec A+1)(\sec A-\tan A)}\\ \\ \Longrightarrow\ \ &\frac{1}{\sec A-\tan A}\\ \\ \\ \Longrightarrow\ \ &\large \textsf{...RHS}\end{aligned}$

$\huge \textsc{\underline{\underline{Hence Proved!!!}}}$