Question

Prove that

1/secA-tanA - 1/cosA = 1/cosA- 1/secA+tanA

Answer 1

Answer:

Step-by-step explanation:

Consider 1/secA-tanA - 1/cosA

Multiplying by secA + tanA in the numerator and denominator of

first term,

we get secA + tanA/(secA +tanA)(secA - tanA) - 1/cosA

= secA + tanA - secA (Since sec²A - tan²A = 1)

= tanA

Adding and subtracting secA , we get

secA + tanA - secA

= 1/cosA - (secA - tanA)

Now multiplying and dividing (secA - tanA) by (secA + tanA), we

get 1/cosA - (sec²A - tan²A)/(secA + tanA)

= 1/ cosA - 1/secA + tanA

= R.H.S

Answer 2

$\boxed{Explained\:Answer}$

______________________________

Consider 1/secA-tanA - 1/cosA

Multiply secA + tanA in the numerator and denominator of first term,

we get secA + tanA/(secA +tanA)(secA - tanA) - 1/cosA

= secA + tanA - secA (Since sec²A - tan²A = 1)

= tanA

Adding and subtracting secA , we get

secA + tanA - secA

= 1/cosA - (secA - tanA)

Now multiplying and dividing (secA - tanA) by (secA + tanA), we

get 1/cosA - (sec²A - tan²A)/(secA + tanA)

= 1/ cosA - 1/secA + tanA

= R.H.S

Hence, Proved.