prove that
cos(90-theta)*sin(90-theta)/tan(90-theta)=sin^theta
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Solution :
Here I am using A instead of theta.
LHS = [cos(90-A)sin(90-A)]/tan(90-A)
= ( sinAcosA)/cotA
= ( sinAcosA )/( cosA/sinA )
= ( sinAcosA × sinA )/cosA
After cancellation, we get
= sin²A
= RHS
••••
Here I am using A instead of theta.
LHS = [cos(90-A)sin(90-A)]/tan(90-A)
= ( sinAcosA)/cotA
= ( sinAcosA )/( cosA/sinA )
= ( sinAcosA × sinA )/cosA
After cancellation, we get
= sin²A
= RHS
••••
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2
...............cos(90+x)=-sinx, where 90 and x are both measured in degrees.
L.H.S. = cos(90+x)= cos(180–90+x)=cos[180–(90-x)]=-cos(90-x)=-sin x
This transformation process makes use of two trigonometric identities,
cos(180-x)=-cosxcos(90-x)=sinx
There are also similar identities, and they can be conveniently remembered by using the concept of quadrants (“CAST”):
L.H.S. = cos(90+x)= cos(180–90+x)=cos[180–(90-x)]=-cos(90-x)=-sin x
This transformation process makes use of two trigonometric identities,
cos(180-x)=-cosxcos(90-x)=sinx
There are also similar identities, and they can be conveniently remembered by using the concept of quadrants (“CAST”):
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