Question

tantita +sec -1/ tantita - sec +1 =sin+1/cos

Answer 1

Hey there !!

Prove that :-

→ $\bf \frac{ sec \theta + tan \theta - 1 }{ tan \theta - sec \theta + 1 } = \frac{ sin \theta + 1 }{ cos \theta } .$

Solution :-

[ Solving LHS ]

→ $\bf \frac{ sec \theta + tan \theta - 1 }{ tan \theta - sec \theta + 1 } = \frac{ sin \theta + 1 }{ cos \theta } .$

=  $\bf \frac{ ( sec \theta + tan \theta ) - ( {sec}^{2} - {tan}^{2} ) }{ tan \theta - sec \theta + 1 } .$

[ ∵ sec²θ - tan²θ = 1 ]

=  $\bf \frac{ ( sec \theta + tan \theta ) - [ 1 - ( sec \theta - tan \theta ) }{ tan \theta - sec \theta + 1 } .$

=  $\bf \frac{ ( sec \theta + tan \theta ) - \cancel{( tan \theta - sec \theta + 1 ) }}{ \cancel{ ( tan \theta - sec \theta + 1 ) }} .$

= sec θ + tan θ .

= $\bf \frac{1}{cos \theta } + \frac{sin \theta }{cos \theta } .$

= $\boxed{ \bf \frac{1 + sin \theta }{cos \theta } . }$

LHS = RHS.

Hence, it is proved.

THANKS.

#BeBrainly.

Answer 2

Step-by-step explanation:

follow me

And Hope it helps u