Question

If tan A + cot A = √5 then find tan^3A + cot^3 A

Answer 1

Answer:
${tan}^{3}A + {cot}^{3}A= 2 \sqrt{5}$

Solution:

$tan A+cot A=\sqrt{5}\\\\tan \: A + \frac{1}{tan \: A} = \sqrt{5} \\ \\ {tan}^{2} A - \sqrt{5} tan \: A + 1 = 0$
by Quadratic formula

$tan \: A_{1,2} = \frac{ \sqrt{5} ±\sqrt{5 - 4} }{2} \\ \\ tan A = \frac{ \sqrt{5} + 1}{2} \\ \\ tan \: A = \frac{ \sqrt{5} - 1}{2}$

Since

${x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} ) \\ \\ {tan}^{3}A + {cot}^{3}A= (tan \: A + cot \: A)( {tan}^{2}A - tan \: A. cot \: A + {cot}^{2} A) \\ \\ = \sqrt{5} (( { \frac{ \sqrt{5} + 1 }{2} })^{2} - 1 + ( { \frac{2}{ \sqrt{5} + 1} })^{2} ) \\ \\ = \sqrt{5} ( \frac{5 + 1 + 2 \sqrt{5} }{4} - 1 + \frac{4}{5 + 1 + 2 \sqrt{5} } ) \\ \\ \sqrt{5} ( \frac{6+ 2 \sqrt{5} }{4} - 1 + \frac{4}{6+ 2 \sqrt{5} } ) \\ \\ = \sqrt{5} ( \frac{3 + \sqrt{5} }{2} - 1 + \frac{2}{3 + \sqrt{5} } ) \\ \\ = \sqrt{5} ( \frac{9 + 5 + 6 \sqrt{5} - 6 - 2 \sqrt{5} + 4 }{2(3 + \sqrt{5} )} ) \\ \\ = \sqrt{5} ( \frac{12 + 4 \sqrt{5} }{2(3 + \sqrt{5}) }) \\ \\ = \sqrt{5} ( \frac{6 + 2 \sqrt{5} }{3 + \sqrt{5} } ) \\ \\ {tan}^{3}A + {cot}^{3}A= 2 \sqrt{5}$

This can be solved by another value of tan A.

Answer 2

HELLO DEAR,


we know:- (a³ + b³) = a³ + b³ + 3ab(a + b)

now,

GIVEN:- tanA + cotA = √5

so, on cubing both side, we get

tan³A + cot³A + 3tanA . cotA(tanA + cotA) = 5√5

=> tan³A + cot³A + 3(√5) = 5√5

=> (tan³A + cot³A) = 5√5 - 3√5

=> (tan³A + cot³A) = 2√5


I HOPE IT'S HELP YOU DEAR,
THANKS