Question

SecA+tanA/cosecA+cotA=cosecA-cotA/secA-tanA

Answer 1

Answer:

Step-by-step explanation:

we know that

$sec^{2}A-tan^{2}A=1\\\\(secA+tanA)(secA-tanA)=1\\\\secA+tanA=\frac{1}{secA-tanA}.......(1)$

we know that

$cosec^{2}A-cot^{A}=1\\\\(cosecA+cotA)(cosecA-cotA)=1\\\\cosecA+cotA=\frac{1}{cosecA-cotA}....(2)$

$Now,\\\frac{secA+tanA}{cosecA+cotA}$

by using equations (1)&(2)

$=\frac{\frac{1}{secA-tanA}}{\frac{1}{cosecA-cotA}}$

$=(\frac{1}{secA-tanA}).(\frac{cosecA-cotA}{1})\\\\=\frac{cosecA-cotA}{secA-tanA}$

I hope this answer helps you

Answer 2

Answer:

Hope it helps you friend.