Question

secA(1-sinA)(secA+tana)=1 prove that ​

Answer 1

Answer:

SecA ( 1 - sinA ) (secA + tan A)  

= (secA - sinA × secA)( secA + tanA )  

= (secA - sinA/cosA)( secA + tanA )

= ( secA - tanA) ( secA + tanA)

= sec²A - tan²A

=  1

Answer 2

To Prove:

• $secA ( 1 - sinA )(secA + tanA ) = 1$

Step-by-step explanation:

Given,

$L.H.S\\\\ = sec A ( 1 - sin A ) ( sec A + tan A ) \\\\ = ( \sec A - \sin A \sec A)( \sec A + \tan A)$

But,

we know that,

$\sec A = \frac{1}{ \cos A}$

Thus,

pUtting the values,

we get,

$= ( \sec A - \frac{ \sin A}{ \cos A} )( \sec A + \tan A)$

But,

we know that,

$\frac{ \sin \: A }{ \cos \: A } = \tan A$

Thus,

pUtting the values,

we get,

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

Now,

we know that,

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

Thus,

we get,

$= { \sec }^{2} A - { \tan }^{2} A \: \\ \\ = 1 \\ \\ = \: R.H.S$

Thus,

L.H.S = R.H.S

Hence, Proved