Question

if 4cota=3 then sinA +cosA / sinA-cosA

Answer 1

Answer: 7 mark as brainlist

Step-by-step explanation:

cot A =3/4 i.e b= 3 and p = 4

then from pythagoras theorem,

h= $\sqrt{p^2 +b^2$ = $\sqrt{4^2+3^2$ = 5

then sin A = p/h = 4/5

cos A = b/h = 3/5

Then,

(sinA +cosA) / (sinA-cosA) = (4/5 + 3/5 )  /  (4/5 - 3/5 ) = 7

Answer 2

Answer:

The answer is 7

Step-by-step explanation:

Let us take a right angled triangle ABC where B=90 degrees, AB= 4cm and BC=3cm and AC=5 cm (caluclations based on pythogoras theorem).  

cotA = 3/4 (adj/opp)

So sinA = (opp/hypo) = 4/5. Similarly cosA = 3/5.

Now, sinA + cosA/ sinA- cosA = 4/5 + 3/5 / 4/5 - 3/5.

Further caluclating, we will get  7/5 / 1/5. If we take 1/5 to the numerator side then it will become 7/5 * 5.

5 and 5 will get cancelled.

so the answer is 7.

hope it helped you...