is -tan(b-a)=tan(a-b)
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Answered by
0
yes... of course -tan(b-a)=tan(a-b)
-tan(b-a)=tan(a-b)
LHS=-tan(b-a)
=-tan[-(a-b)]
=-[-tan(a-b)]
=tan(a-b)
{because, tan-(x)= -tanx}
=RHS
therefore, -tan(b-a)=tan(a-b)
-tan(b-a)=tan(a-b)
LHS=-tan(b-a)
=-tan[-(a-b)]
=-[-tan(a-b)]
=tan(a-b)
{because, tan-(x)= -tanx}
=RHS
therefore, -tan(b-a)=tan(a-b)
Answered by
1
lhs = - tan (b-a)
= - tan [-(a-b)]
= - [- tan( a- b )]
[ since tan ( - θ ) = - tan θ ]
= tan ( a - b )
= RHS
hope it will helps you.
= - tan [-(a-b)]
= - [- tan( a- b )]
[ since tan ( - θ ) = - tan θ ]
= tan ( a - b )
= RHS
hope it will helps you.
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