Question

sec A + tan A -1 / tan A - secA = 1+ sin A / cosA​

Answer 1

Correct QuestioN :

$\tt : \implies \dfrac{Sec A + tan A - 1 }{ tan A - Sec A + 1 } =\dfrac{1 + Sin A}{Cos A}$

SolutioN :

★ Taking LHS,

$\tt : \implies \dfrac{Sec A + tan A - 1 }{ tan A - Sec A + 1 }$

We know,

$\tt \: 1 + tan^{2} A = sec^2 A. \\ \tt \: 1 = sec^2 A - {tan}^{2} A.$

★ Substitute as place of - 1, We get.

$\tt : \implies \dfrac{Sec A + tan A - \Big({Sec }^{2} A - { tan }^{2} A \Big)}{ tan A - Sec A + 1 }$

We know,

• a² - b² = ( a + b )( a - b )

$\tt: \implies \dfrac{Sec A + tan A - \Big(Sec A + tan A \Big)\Big(Sec A - tan A\Big)}{ tan A - Sec A + 1 }$

★ Taking Common ( Sec A + tan A. )

$\tt: \implies \dfrac{Sec A + tan A \Big[1- \Big(Sec A - tan A\Big)\Big]}{ tan A - Sec A + 1 }$

$\tt: \implies \dfrac{Sec A + tan A \cancel{\Big(1- Sec A + tan A\Big)}}{\cancel{ tan A - Sec A + 1} }$

$\tt: \implies Sec A + tan A$

$\tt : \implies \dfrac{1}{Cos A} + \dfrac{Sin A}{Cos A}$

$\tt : \implies \dfrac{1 + Sin A}{Cos A}$

Hence Proved.