Math, asked by ayushbose620, 1 year ago

cosx+secx=2. find cos^5 x+sin^9 x​

Answers

Answered by rishu6845
0

Answer:

1

Step-by-step explanation:

Given---> Cosx + Secx = 2

To find ---> Cos⁵x + Sin⁹x = ?

Solution---> ATQ,

Cosx + Secx = 2

We know that Secx = 1 / Cosx , applying it here,

=> Cosx + ( 1 / Cosx ) = 2

Taking Cosx as LCM , we get

=> ( Cos²x + 1 ) / Cosx = 2

=> Cos²x + 1 = 2 Cosx

=> Cos²x - 2Cosx + 1 = 0

We have an identity ,

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

=> ( Cosx - 1 )² = 0

Taking square root of both sides we get,

=> Cosx - 1 = 0

=> Cosx = 0

=> Cosx = Cos0°

=> x = 0°

Now, Cos⁵x + Sin⁹x = ( Cosx )⁵ + ( Sinx )⁹

Putting x = 0° in it , we get

= ( Cos0° )⁵ + ( Sin0° )⁹

= ( 1 )⁵ + ( 0 )⁹

Cos⁵x + Sin⁹x = 1 + 0

Cos⁵x + Sin⁹x = 1

Answered by Aɾꜱɦ
17

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1

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