Question

if sin A= 3/5 then find the following
(sec2)A+ (tan2)A

Answer 1

Answer:

cosec A = 5/3

cos A = √1- sin² = √1-(3/5) ²= √1-9/25=4/5

sec A = 5/4

tan A = sin A/ cos A = 3/4

therefore, (5/4) ²A+ (3/4) ² A = 25/16 + 9/16= 34/16 = 17/8

Step-by-step explanation:

i don't know if this is correct but I hope you understand the steps for it

Answer 2

Step-by-step explanation:

Given, sinA=53=hp

∴b=25−9=16=4

secA=bh=45,

tanA=bp=43

Therefore, sec2A−tan2A=(45)2−(43)2

=1625−169=1616=1