Question

prove that tan A -1+sec A / tan A +1 -sec A = sec a + tan A

Answer 1

Answer:

To Prove:

\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}tanA−secA+1tanA+secA−1=cosA1+sinA

Solution:    

\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{\tan A+\sec A-\left(\sec ^{2} A-\tan ^{2} A\right)}{\tan A-\sec A+1}tanA−secA+1tanA+secA−1=tanA−secA+1tanA+secA−(sec2A−tan2A)

\left[\because 1+\tan ^{2} A=\sec ^{2} A \Rightarrow \sec ^{2} A-\tan ^{2} A=1\right][∵1+tan2A=sec2A⇒sec2A−tan2A=1]

=\frac{\tan A+\sec A-\{(\sec A-\tan A)(\sec A+\tan A)\}}{\tan A-\sec A+1}=tanA−secA+1tanA+secA−{(secA−tanA)(secA+tanA)}

Now, take(\tan A+\sec A)(tanA+secA) as a common term, we get

=\frac{(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}=tanA−secA+1(tanA+secA)(1−secA+tanA)

=\frac{(\tan A+\sec A)(\tan A-\sec A+1)}{\tan A-\sec A+1}=tanA−secA+1(tanA+secA)(tanA−secA+1)

=\tan A+\sec A=tanA+secA

=\frac{\sin A}{\cos A}+\frac{1}{\cos A}=cosAsinA+cosA1    

=\frac{1+\sin A}{\cos A}=cosA1+sinA    

∴ Hence Proved

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