Question

If Cot∅ + cos∅ = p , Cot∅ - Cos∅ = q , prove that ( p² - q² )² = 16pq ??

Answer 1

p = cot∅ + cos∅
q = cot∅ - cos∅ , is given

( p² - q² )² = 16pq

on putting value
[( cot∅ + cos∅)² - ( cot∅ - cos∅)²]²

[ cot²∅ + cos²∅ + 2cot∅×cos∅-( cot²∅ + cos²∅ - 2cot∅×cos∅)]²

[ cot²∅+ cos²∅+2cot∅×cos∅-cot²∅ - cos²∅ + 2cot∅cos∅)²

( 4cot∅cos∅)²
16cot²∅× cos²∅
16

Hence proved

wohoooooooo......

Answer 2

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 /