Question

cot 15. Prove that: cotA/1+tanA=cotA-1/2-sec2A
​

Answer 1

Answer:

Notice,

LHS=cotA−12−sec2ALHS=cot⁡A−12−sec2⁡A

=1tanA−12−(1+tan2A)=1tan⁡A−12−(1+tan2⁡A)

=1−tanAtanA1−tan2A=1−tan⁡Atan⁡A1−tan2⁡A

=1−tanAtanA(1−tan2A)=1−tan⁡Atan⁡A(1−tan2⁡A)

=1−tanAtanA(1−tanA)(1+tanA)=1−tan⁡Atan⁡A(1−tan⁡A)(1+tan⁡A)

=1tanA(1+tanA)=1tan⁡A(1+tan⁡A)

=cotA1+tanA=cot⁡A1+tan⁡A

=RHS

mark me brianlist if it helps

Answer 2

Step-by-step explanation:

Here is your answer dear