Question

Prove that:
cotQ-1/2-sec2Q=cotQ/1+tanQ​

Answer 1

Step-by-step explanation:

Given:-

(Cot Q -1)/(2-Sec^2Q)

To Prove :-

(Cot Q -1)/(2-Sec^2Q) = CotQ/(1+tanQ)

Proof:-

LHS :-

(Cot Q -1)/(2-Sec^2 Q)

=> (Cot Q -1)/(1+1-Sec^2 Q)

We know that

Sec^2 A - Tan^2 A = 1

=> (Cot Q -1)/ (1-Tan^2 Q)

=> [(1/Tan Q)-1]/[(1+Tan Q)(1-Tan Q)]

Since (a+b)(a-b)=a^2-b^2

=> [(1-Tan Q)/Tan Q ]/[(1+Tan Q)(1-Tan Q)]

On cancelling (1- Tan Q)

=> (1/TanQ )/(1+Tan Q)

=> Cot Q /(1+Tan Q)

Since Cot A = 1/Tan A

=> RHS

LHS = RHS

Hence, Proved

Used formulae:-

• Sec^2 A - Tan^2 A = 1

• Cot A = 1/Tan A

• (a+b)(a-b)=a^2-b^2