Question

If tan theta + cot theta = 3, find
the value of (tan^2 theta + cot^2 theta)(tan^3 theta + cot^3 theta).

please answer this question and help me ....
thanks in advance​

Answer 1

Step-by-step explanation:

tanO + CotO = 3 -----(1)

Doing square both side

tan²O + Cot²O + 2 = 9

tan²O + Cot²O = 7 ----(2)

Cube of equation (1)

tan³O + Cot³O + 3(3) = 27

tan³O + Cot³O = 18 ----(3)

equation (2) × equation (3)

(tan²O+Cot²O)(tan³O+Cot³O) = 7*18 = 126

Answer 2

HEY MATE YOUR ANSWER IS HERE...

(TAKING THETA AS “ X ” )

$\tan(x) + \cot(x) \: = 3$

SQUARING BOTH SIDES ..

$( \tan(x) + \cot(x) ) ^{2} = 9$

${ \tan }^{2} (x) + { \cot(x) }^{2} + 2 \tan(x) \cot(x) = 9$

NOW TAN THETA × COT THETA = 1

${ \tan }^{2}(x) + \ { \cot }^{2}( x) = 9 - 2$

${ \tan }^{2}(x) + \ { \cot }^{2}( x) = 7 \: \: be \: eq \: 1$

NOW IN CASE 2

$\tan(x) + \cot(x) \: = 3 \:$

NOW CUBING BOTH SIDES

${(\tan(x) + \: \cot(x) )}^{3} = 27$

${ \tan }^{3} x \: + { \cot }^{3} x \: + 3 \tan(x) \cot(x) ( \tan(x) + \cot(x) ) = 27$

NOW AGAIN TAN THETA × COT THETA = 1

AND TAN THETA + COT THETA = 3 ( GIVEN)

${ \tan }^{3} x \: + { \cot }^{3} x + 3(3) = 27$

${ \tan }^{3} x \: + { \cot }^{3} x = 27 \: - 9$

${ \tan }^{3} x \: + { \cot }^{3} x = 18 \: be \: eq \: 2$

★ NOW FROM EQ 1 AND 2 ★

PUTTING THEIR VALUES

$( \: \: { \tan}^{2} (x) \: + { \cot}^{2} (x) \: \: )( \: \: { \tan }^{3} (x) \: \: + { \cot }^{3} (x) \: \: )$

$= (7)(18)$

$= 126$

THANKS FOR YOUR QUESTION HOPE THIS HELPS

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