Question

prove.. tan^2A/tanA-1+cotA/1-tanA=1+secAcosecA​

Answer 1

Step-by-step explanation:

Solution :-

On taking LHS :

[Tan² A / (Tan A -1)] + [ Cot A / (1-Tan A]

=>[Tan² A /(Tan A -1)] + [ Cot A / {-(Tan A-1)}]

=> [Tan² A / (Tan A -1)] - [ Cot A / (Tan A-1)]

=> (Tan² A - Cot A )/(Tan A - 1)

=> [Tan² A - (1/Tan A )] / (Tan A - 1)

Since Cot A = 1/Tan A

=>[ (Tan³ A-1)/Tan A ]/(Tan A -1)

=> (Tan³ A -1)/[Tan A (Tan A-1)]

=> (Tan³ A -1³)/[Tan A (Tan A-1)]

=> [TanA-1)(Tan²A+(Tan A)(1)+1²)]/

[Tan A(Tan A-1)]

Since a³-b³ = (a-b)(a²+ab+b²)

=>[TanA-1)(Tan²A+TanA+1)]/[TanA(TanA-1)]

On cancelling Tan A -1

=> (Tan² A + Tan A +1)/ Tan A

=>(Tan²A/Tan A )+(Tan A/Tan A ) +(1/Tan A)

=> Tan A + 1 + Cot A

=> (Sin A / Cos A ) + 1 + (Cos A / Sin A )

Since , Tan A = Sin A / Cos A

Cot A = Cos A / Sin A

=>(Sin²A+SinACosA+Cos²A)/(SinACosA )

=> (1+Sin A Cos A )/(Sin A Cos A )

Since Sin² A + Cos² A = 1

=>(1/SinACosA )+(SinACosA/SinACosA )

=> (1/Sin A Cos A )+1

=> (1/Sin A ) (1/Cos A ) +1

We know that

1/Sin A = Cosec A

1/Cos A = Sec A

=> Cosec A Sec A +1

=> 1+ Cosec A Sec A

=> 1+ Sec A Cosec A

=> RHS

=> LHS = RHS

Hence, Proved.

Used formulae:-

→ a³-b³ = (a-b)(a²+ab+b²)

→ Tan A = Sin A / Cos A

→ Cot A = Cos A / Sin A

→ Sin² A + Cos² A = 1

→ 1/Sin A = Cosec A

→ 1/Cos A = Sec A

→ a×b = b×a