Question

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$if \: tan \alpha + cot \alpha = 4 \: then \: {tan}^{4} \alpha + {cot}^{4} \alpha =$
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Answer 1

Answer:

194

Step-by-step explanation:

Given---> tanα + Cotα = 4

To find---> tan⁴α + Cot⁴α = ?

Solution---> ATQ,

tanα + Cotα = 4

Squaring both sides we get,

=>( tanα + Cotα )² = ( 4 )²

We know that,

( a + b )² = a² + b² + 2ab , applying it we get,

=> ( tanα )² + ( Cotα )² + 2 tanα Cotα = 16

We know that Cotα = 1 / tanα , applying it , we get,

=> tan²α + Cot²α + 2 tanα ( 1 / tanα ) = 16

=> tan²α + Cot²α + 2 = 16

=> tan²α + Cot²α = 16 - 2

=> tan²α + Cot²α = 14

Squaring both sides again , we get

=> ( tan²α + Cot²α )² = ( 14 )²

=> ( tan²α )² + ( Cot²α )² + 2 tan²α Cot²α = 196

We know that, tan²α = 1 / Cot²α , applying it we get,

=> tan⁴α + Cot⁴α + 2 tan²α ( 1 / tan²α ) = 196

=> tan⁴α + Cot⁴α + 2 = 196

=> tan⁴α + Cot⁴α = 196 - 2

=> tan⁴α + Cot⁴α = 194

Answer 2

Answer:

Step-by-step explanation:

tanα + Cotα = 4

tan⁴α + Cot⁴α = ?

Squaring both sides we get,

=>( tanα + Cotα )² = ( 4 )²

We know that,

( a + b )² = a² + b² + 2ab ,

=> ( tanα )² + ( Cotα )² + 2 tanα Cotα = 16

We know that Cotα = 1 / tanα ,

=> tan²α + Cot²α + 2 tanα ( 1 / tanα ) = 16

=> tan²α + Cot²α + 2 = 16

=> tan²α + Cot²α = 16 - 2

=> tan²α + Cot²α = 14

Squaring both sides again , we get

=> ( tan²α + Cot²α )² = ( 14 )²

=> ( tan²α )² + ( Cot²α )² + 2 tan²α Cot²α = 196

We know that, tan²α = 1 / Cot²α ,

=> tan⁴α + Cot⁴α + 2 tan²α ( 1 / tan²α ) = 196

=> tan⁴α + Cot⁴α + 2 = 196

=> tan⁴α + Cot⁴α = 196 - 2

=> tan⁴α + Cot⁴α = 194