prove the trigonometric identities
Attachments:
Answers
Answered by
4
LHS
= tan²∅ - sin²∅
= sin²∅/cos²∅ - sin²∅
= sin²∅- sin²∅cos²∅/cos²∅
= sin²∅(1-cos²∅)/cos²∅
= sin²∅(sin²∅)/cos²∅
= sin²∅/cos²∅ × sin²∅
= tan²∅ × sin²∅
LHS = RHS
= tan²∅ - sin²∅
= sin²∅/cos²∅ - sin²∅
= sin²∅- sin²∅cos²∅/cos²∅
= sin²∅(1-cos²∅)/cos²∅
= sin²∅(sin²∅)/cos²∅
= sin²∅/cos²∅ × sin²∅
= tan²∅ × sin²∅
LHS = RHS
marakroma18otss61:
how you got tan2 = sin/cos...im confuse in this part
no bro we know tan∅ = sin∅/cos∅ then square both side tan²∅ = sin²∅/cos²∅
bro see once again i write tan²∅ = sin²∅/cos²∅ not tan²∅ = sin∅/cos∅. //thank you//
thank you...
but bro...in trigonometric identities tan^2@= sec^2@-1
yes it is also correct tan²∅ = sec²∅-1 ,. tan²∅ = 1/cos²∅ - 1 ,. tan²∅ = 1-cos²∅/cos²∅,. tan²∅ = sin²∅/cos²∅
thank you very much...
Answered by
3
Hey friend
Here is your answer
TO PROVE :
tan²@ - sin²@= tan²@sin²@
LHS:
=tan²@ - sin²@
=(sin²@/cos²@)-sin²@
=sin²@ [ (1/cos²@)-1 ]
=sin²@ [ (1-cos²@)/cos²@ ]
[USING sin²@+cos²@=1
1-cos²@=sin²@]
=sin²@(sin²@/cos²@)
[USING sin@/cos@=tan@]
=sin²@tan²@
=RHS
HENCE PROVED
==================================
HOPE THIS HELPS YOU
Here is your answer
TO PROVE :
tan²@ - sin²@= tan²@sin²@
LHS:
=tan²@ - sin²@
=(sin²@/cos²@)-sin²@
=sin²@ [ (1/cos²@)-1 ]
=sin²@ [ (1-cos²@)/cos²@ ]
[USING sin²@+cos²@=1
1-cos²@=sin²@]
=sin²@(sin²@/cos²@)
[USING sin@/cos@=tan@]
=sin²@tan²@
=RHS
HENCE PROVED
==================================
HOPE THIS HELPS YOU
Similar questions