Question

prove that :
tanA/1+secA - tanA/1-secA =2cosec​

Answer 1

Answer:

L. H. S.

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

Taking LCM of the denominators,

= tan A ( 1 - sec A) - tan A ( 1 + sec A) /

( 1 + sec A ) ( 1 - sec A)

= tan A ( - 1 sec A - 1 - sec A) / 1 - sec ^2 A

( : , ( 1 + sec A) ( 1 - sec A) = 1 - sec ^2 A

= tan A ( - 2 sec A / 1 - sec ^2

= 2 tan A . sec A / sec ^2 A - 1

= 2 tan A . sec A / tan ^2 A

(: sec ^2 A - tan ^2 A = 1 => sec ^2 A - 1 = tan^2A )

= 2 sec^2 A / tan ^2 A = 2 cos A / cos A sin A

(: sec A = 1/cos A and tan A = sin A / cos A )

= 2/sin A

= 2 cosec A (: 1/sinA = cosec A)

Therefore, RHS .

Hence proved....

Step-by-step explanation:

@MNF

Answer 2

Answer:

1+secA

tanA

−

1−secA

tanA

=

2⋅cosecA

L.H.S

1+secA

tanA

−

1−secA

tanA

=

1−sec

2

A

tanA(1−secA)−tanA(1+secA)

=

−tan

2

A

tanA(1−secA−1−secA)

=

−tanA

−2secA

=

sinA/cosA

2⋅1/cosA

=

sinA

2

= 2⋅ cosecA

=R.H.S

∴ L.H.S= R.H.S.