Question

Prove that:
SecA+1+tanA/SecA+1+tanA=secA+tanA

Answer 1

Answer:

To Prove:

\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}

tanA−secA+1

tanA+secA−1

=

cosA

1+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+1

tanA+secA−1

=

tanA−secA+1

tanA+secA−(sec

2

A−tan

2

A)

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

2

A=sec

2

A⇒sec

2

A−tan

2

A=1]

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

tanA−secA+1

tanA+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}=

cosA

sinA

+

cosA

1

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

cosA

1+sinA

∴ Hence Proved