Question

sec A-tan A=a+1/a-1 theb cos A​

Answer 1

Answer:

Sec^2@-tan^2@=1

or (sec@+tan@)(sec@-tan@)=1

or (sec@+tan@)(a+1)/(a-1)=1

or (sec@+tan@)=(a-1)/(a+1)..... (1)

and (sec@-tan@)= (a+1)/(a-1).... (2)

adding (1) n (2)..

2sec@=[(a-1)^2+(a+1)^2]/(a^2-1)=2 (a^2+1)/(a^2-1)

or sec@=(a^2+1)/(a^2-1)