Question

first answer will be marked as brainlist​

Answer 1

ANSWER:-

Draw a right angle triangle ΔABC

right angle at A

as shown in the adjoining figure in above pic

as we know given

cot (a)=4/3

let AB =4k

AC = 3k

where K is some positive real number..

now by using Pythagoras theorem

BC² =AB² +AC²

put the value we get

BC² = (4K)² +(3K)²

= 16 K² +9 K²

= 25K²

taking squre root out we get..

BC = 5k

Now

tan(a)=ac/ab=3k/4k=3/4

sin(a)=ac/bc=3k/5k=3/5

cos(a)=ab/bc= 4k/5k=4/5

now as we have

1-tan²a/1+tan²a=cos²a-sin²a

take L.H.S AND R.H.S

LHS =

1-tan²a/1+tan²a

putting all the value we get

1-(3/4)²/1+(3/4)²

1-9/16/1+9/16

16-9/16/16+9/16

7/16×16/25 [reciprocal]

L.HS =7/25

now take RHS

cos²(a)-sin²(a)

putting the value we get

1²-(3/5)²

1-9/25

25-9/25

LHS=7/5

there for

we get LHS = RHS

Hope u are satisfied...

please mark me as the BRAINLIEST...✌️