Question

sec(y)sin(π2−y)

Answer 1

secy*sin(2pi-y)=(1/cosy)*siny=siny/cosy=tany

Answer 2

\[\sec y \sin(pie/2-y)\]

Answer 3

this the question

Answer 4

secy=1/cosy
sin(2pi-y)=sin(2pi)cosy-cos(2pi)siny....
sin(2pi)=0...we know that
cos(2pi)=-1 we know that
plug all the values you will get the result as tany

Answer 5

here secy*sin(pi/2-y)
we can rewrite as...(1/cosy).cosy=1

Answer 6

sin(pi/2-y)=sin(pi/2)cosy-cos(pi/2)siny
sin(pi/2)=1
cos(pi/2)=0
so...after plugging these values we will get..
(1/cosy)*cosy=cosy/cosy=1

Answer 7

how did u get cos for sin(pi/2) iam confused in that

Answer 8

Here, we know that secy=1/cosy
so our equation will become sin(pi/2-y)/cosy..
Now, we have the formula for sin(a-b) as..sina*cosb-cosa*sinb
applying that formula we will get our equation as...
sin(pi/2-y)/cosy=(sin(pi/2)cosy-cos(pi/2)siny)/cosy....
we have the defined values of sin(pi/2)=1 cos(pi/2)=0..
so, after plugging that value we will get..
(cosy-0)/cosy=cosy/cosy=1

Answer 9

Let me know if u got it?

Answer 10

ya i got u thx

Answer 11

you r welcome:)