1| 52. Prove that oL तर] secA-tanA cosA. 1B Sec A SecAtian A [CBSE

1.	1\| 52. Prove that oL तर] secA-tanA cosA. 1B Sec A SecAtian A [CBSE 2016]
Answer» 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 Hence, Proved.

1| 52. Prove that oL तर] secA-tanA cosA. 1B Sec A SecAtian A [CBSE 2016]

Answer»

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

Hence, Proved.