Question

$\frac{{tan}^{3} - 1 }{tan - 1 } = {sec}^{2} + tan$
prove
LHS=RHS​

Answer 1

To prove---->

( tan³θ - 1 ) / ( tanθ - 1 ) = Sec²θ + tanθ

Proof -----> First , we solve ( tan³θ + 1 )

We know that,

( a³ - b³ ) = ( a - b ) ( a² + b² + ab ) , applying it we get,

( tan³θ - 1 ) = ( tanθ )³ - ( 1 )³

= ( tanθ - 1 ) { ( tanθ )² + ( 1 )² + ( tanθ ) ( 1 ) }

= ( tanθ - 1 ) ( tan²θ + 1 + tanθ )

We know that , ( 1 + tan²θ ) = Sec²θ , applying it here , we get ,

= ( tanθ - 1 ) ( Sec²θ + tanθ )

Now , taking , LHS

= ( tan³θ - 1 ) / ( tanθ - 1 )

Now, putting value of ( tan³θ - 1 ) , we get,

= ( tanθ - 1 ) ( Sec²θ + tanθ ) / ( tanθ - 1 )

Cancelling ( tanθ - 1 ) , from , numerator and denominator , we get,

= ( Sec²θ + tanθ ) = RHS