cosx+secx=2. find cos^5 x+sin^9 x
Answers
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
1
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