Question

Prove that sinA-cosA+1/sinA-cosA-1=1/secA-tanA using sec²A=1+tan²A​

Answer 1

Step-by-step explanation:

LHS...

sinA+cosA−1

sinA−cosA+1

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

\begin{gathered}\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}\end{gathered}

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sinA+cosA−1

sinA−cosA+1

(sinA+cosA−1)secA

(sinA−cosA+1)secA

sinA⋅secA+cosA⋅secA−secA

sinA⋅secA−cosA⋅secA+secA

tanA+1−secA

tanA−1+secA

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

Now, multiplying both the numerator and the denominator with

(secA−tanA),

\begin{gathered}\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}\end{gathered}

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tanA+1−secA

tanA−1+secA

(tanA−secA+1)(secA−tanA)

(tanA+secA−1)(secA−tanA)

\begin{gathered}\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}\end{gathered}

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(tanA−secA+1)(secA−tanA)

sec

2

A−tan

2

A−secA+tanA

(tanA−secA+1)(secA−tanA)

1+tan

2

A−tan

2

A−secA+tanA

[∵ sec

2

A=1+tan

2

A]

(tanA−secA+1)(secA−tanA)

tanA−secA+1

secA−tanA

1

...RHS

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

HenceProved!!!