Question

cotA +cos A = p
cotA - cos A = q

(1) prove that (p²-q²)² = 16 p

(2)prove that ,
cotA = (2√(pq)) ÷ (p-q)​

Answer 1

Answer:

Step-by-step explanation:

Given,

cot∅ + cos∅ = P

cot∅ - cos∅ = q

add both equation ,

2cot∅ = (P + q)

cot∅ = (P + q)/2 --------(1)

again, subtract first to second equation

2cos∅ = (P - q)

cos∅ = (P - q)/2

so, cos∅= base/hypotenuse

so, perpendicular = √(4 - (P - q)² }

so, tan∅ = √{ 4 - (P -q)²}/(p-q) -----(2)

multiply eqns (1) and (2)

tan∅.cot∅ = (P +q)√{4 - (P -q)²}/2(P -q)

1 = (P+ q)√{4-(P-q)²}/2(P-q)

take square both sides,

4(P - q)² =(P + q)²{4 - (P-q)² }

4(P -q)² = 4(P +q)² -{ (P+q)(p-q)}²

4{ P² + q² -2Pq - P² - q² - 2Pq } = -(P² -q²)

- 16Pq = -(P² - q²)

(P² - q²) = 16Pq

hence, proved /