Question

show that : 1/(sec a + tan a) = sec a - tan a​

Answer 1

$\large\underline\mathfrak\red{Given \: :- }$

$\sf{ \frac{1}{ \sec a + \tan a } = \sec a - \tan a}$

$\large\underline\mathfrak\red{To \: Prove \: :- }$

• LHS = RHS

$\large\underline\mathfrak\red{Proof \: :- }$

$\sf{ \frac{1}{ \sec a + \tan a } = RHS}$

$\sf{ \frac{1}{ \sec a + \tan a } \times \frac{\sec a - \tan a}{\sec a - \tan a} =RHS }$

$\sf{1 \times \sec a - \tan a =RHS}$

$\sf{ \sec a - \tan a =RHS}$

$\sf\therefore{LHS = RHS}$

${\large{\underline{\underline{\bf{\red{Hence \: Proved \: ! }}}}}}$

Answer 2

Given,

• 1/(secA + tanA) = secA - tanA

To Proof,

• 1/(secA + tanA) = secA - tanA

Proof,

1/(secA + tanA) = secA - tanA

→ 1/(secA + tanA) × (secA - tanA)/(secA - tanA) = secA - tanA

→ (secA - tanA)/(sec²A - tan²A) = secA - tanA

[sec²A - tan²A = 1]

→ (secA - tanA)/(1) = secA - tanA

→ secA - tanA = secA - tanA

→ L.H.S. = R.H.S.

••••••••••••••••••Hence Proved,