acosθ +bsinθ =p and asinθ-bcosθ =q, then prove that a²+b²=p²+q².Prove it by using trigonometric identities.
Answers
Answered by
1
hope it helps you
if useful plzzz mark as brainliest
if useful plzzz mark as brainliest
Attachments:
Answered by
1
Here, I am using x instead of theta.
It is given that ,
acosx+bsinx = p ----( 1 )
=> (acosx+bsinx)² = p²
=> a²cos²x+b²sin²x+2absinxcosx=p²---(2)
asinx-bcosx=q----(3)
=> (asinx-bcosx)²=q²
=> a²sin²x+b²cos²x-2absinxcosx=q²---(4)
Add (2)&(4) , we get
a²cos²x+a²sin²x+b²sin²x+b²cos²x=p²+q²
=> a²(cos²x+sin²x)+b²(sin²x+cos²x)=p²+q²
=> a²+b²=p²+q²
[ cos²x+sin²x = 1 ]
•••••
It is given that ,
acosx+bsinx = p ----( 1 )
=> (acosx+bsinx)² = p²
=> a²cos²x+b²sin²x+2absinxcosx=p²---(2)
asinx-bcosx=q----(3)
=> (asinx-bcosx)²=q²
=> a²sin²x+b²cos²x-2absinxcosx=q²---(4)
Add (2)&(4) , we get
a²cos²x+a²sin²x+b²sin²x+b²cos²x=p²+q²
=> a²(cos²x+sin²x)+b²(sin²x+cos²x)=p²+q²
=> a²+b²=p²+q²
[ cos²x+sin²x = 1 ]
•••••
Similar questions