Solve this question for Trigonometry Class 10th..
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Answered by
2
Hi ,
***********************************************
As we know the trigonometric identity
sin² A + cos² A = 1
***********************************************
I'm using A instead of theta
LHS =( sin⁴ A + cos⁴ A )/ ( 1 - 2sin²Acos²A)
=[(sin² A)²+(cos² A)²
+2sin²Acos²A-2sin²Acos²A]/
( 1 - 2sin²Acos²A)
= [( sin² A + cos² A)² -2sin²Acos²A]/
(1-2sin²Acos²A)
= ( 1 - 2sin²Acos²A ) / ( 1 - 2sin²Acos²A )
= 1
= RHS
I hope this helps you.
:)
***********************************************
As we know the trigonometric identity
sin² A + cos² A = 1
***********************************************
I'm using A instead of theta
LHS =( sin⁴ A + cos⁴ A )/ ( 1 - 2sin²Acos²A)
=[(sin² A)²+(cos² A)²
+2sin²Acos²A-2sin²Acos²A]/
( 1 - 2sin²Acos²A)
= [( sin² A + cos² A)² -2sin²Acos²A]/
(1-2sin²Acos²A)
= ( 1 - 2sin²Acos²A ) / ( 1 - 2sin²Acos²A )
= 1
= RHS
I hope this helps you.
:)
mysticd:
to make the square we have to add and subtract the same term
i.e + 2sin²Acos²A - 2sin²Acos²A
then first 3terms = ( sin²A + cos² A)²
remaining term is - 2sin²Acos²A
is it clear now . give feed back
:)
Answered by
2
RHS
1- 2sin2teta*cos2teta
=(sin2teta+cos2teta)-2sin2teta*cos2teta
=sin2teta(1-2cos2teta)+cos2teta
=sin2teta(sin2teta+cos2teta-2cos2teta)+cos2teta
=sin2teta(sin2teta-cos2teta)+cos2teta
=sin4teta sin2teta*cos2teta+cos2teta
=sin4teta+cos2teta(1-sin2teta)
=sin4teta+cos2teta(sin2teta+cos2teta-sin2teta)
=sin4teta+cos2teta(+cos2teta)
=sin4teta+cos4teta
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