sec2-cot2(90-theta)=cos2(90-theta)cos2theta
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Hi ,
I think there is an error in the problem.
Here I am using A instead of theta
we know that,
1) cot( 90 - A ) = tanA
2 ) cos ( 90 - A ) = sinA
3 ) sec² A - tan² A = 1
4 ) sin² A + cos² A = 1
Now ,
according to the problem given ,
LHS = sec² A - cot² ( 90 - A )
= sec² A - tan² A
= 1
= sin² A + cos² A
= cos² ( 90 - A ) + cos² A
= RHS
I hope this helps you.
:)
I think there is an error in the problem.
Here I am using A instead of theta
we know that,
1) cot( 90 - A ) = tanA
2 ) cos ( 90 - A ) = sinA
3 ) sec² A - tan² A = 1
4 ) sin² A + cos² A = 1
Now ,
according to the problem given ,
LHS = sec² A - cot² ( 90 - A )
= sec² A - tan² A
= 1
= sin² A + cos² A
= cos² ( 90 - A ) + cos² A
= RHS
I hope this helps you.
:)
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