Question

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

Answer 1

sec A (1 - sin A) (sec A + tan A)

= (sec A - sec A . sin A) (sec A + tan A)

= (sec A - tan A) (sec A + tan A)

= sec²A - tan²A

= 1

Answer 2

Given,

$\[\sec A\left( {1 - \sin A} \right)\left( {\sec A + \tan A} \right) = 1\]$

Solution,

Consider the left hand side of the question,

$\[ = \sec A\left( {1 - \sin A} \right)\left( {\sec A + \tan A} \right)\]$

$\[ = \left( {\sec A - \sec A\sin A} \right)\left( {\sec A + \tan A} \right)\]$

$\[ = \left( {\sec A - \frac{{\sin A}}{{\cos A}}} \right)\left( {\sec A + \tan A} \right)\]$

Know that the formula,$\[\left( {{a^2} + {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\]$

$\[ = \left( {\sec A - \tan A} \right)\left( {\sec A + \tan A} \right)\]$

$\[ = \left( {{{\sec }^2}A - {{\tan }^2}A} \right)\]$

$\[ = 1\]$

Hence result of left hand side is equal to right hand side. So the given problem is proved.