Question

Prove that: sec A (1 – sin A)(sec A + tan A) = 1.​

Answer 1

Given : sec A {1 - sin A} {sec A + tan A} = 1

Proof :

⟹ sec A {1 - sin A} {sec A + tan A} = 1

⟹ sec A {1 - sin A} {1/cos A + sin A/cos A} = 1

⟹ sec A {1 - sin A} {(1 + sin A)/cos A} = 1

⟹ sec A {(1 - sin² A)/cos A} = 1

⟹ sec A {cos²A/cos A} = 1

⟹ sec A {cos A} = 1

⟹ 1/cos A {cos A} = 1

⟹ 1 = 1

∴ It's proved that, sec A {1 - sin A} {sec A + tan A} = 1

{More info : Formulae used in proof are ;

• sec θ = 1/cos θ
• tan θ = sin θ/cos θ
• (a - b) (a + b) = a² - b² i.e. (1 - sin A) (1 + sin A) = (1 - sin² A)
• 1 - sin² θ = cos² θ }

Answer 2

Answer:

$sec \: a (1 - sin \: a \: ) (sec \: a ) (sec \: a + tan \: a) =1$

• Taking LHS

$\frac{1}{cos \: a} (1 - sin \: a)( \frac{1}{cos \: a} + \frac{sin \: a}{cos \: a} )$

$\frac{1 - sin \: a}{cos \: a} \times \frac{1 + sin \: a}{cos \: a}$

$\frac{1 - { sin}^{2}a }{ {cos}^{2}a }$

a ^2 - b^2 = ( a +b ) ( a - b )

$\frac{1 - {sin}^{2}a }{ {cos}^{2}a} \implies \: \frac{ {cos}^{2}a }{ {cos}^{2}a } \implies \: \frac{ \cancel {cos}^{2}a }{ \cancel{cos}^{2}a } \implies \: 1$

$1= 1$

LHS = RHS

Hence proved .