Question

If cosA>0°, tanA+SinA=m, tanA-SinA=n then show that m²-n²=4√mn​

Answer 1

HOPE YOU UNDERSTAND

L.h.s

M²-n²

(m+n)(m-n)

(tanA+sinA+tanA-sinA)(tanA+sinA-tanA+sinA)

(2tanA)(2sinA)

(2sinA/cosA)(2sinA)

4sin²A/cosA

r.h.s

4√mn

4√(tanA+SinA)(tanA-SinA)

4√tan²A-tanA*sinA+sinA*tanA-sin²A

4√tan²A-sin²A

4√sin²A/cos²A-sin²A

4√sin²A-sin²A*cos²A/cos²A

4√sin²A(1-cos²A)/cos²A                      (SINCE 1-COS²A=SIN²A)

4√(sin²A)²/cos²A

4sin²A/cosA

THEREFORE L.H.S=R.H.S PROVED

IF YOU WANT TRIGNOMETRY ANSWER THEN ASK ME IN MESSANGER

Answer 2

Answer:

your answer attached in the photo

you can replace θ by A