Question

ab +a-b+ 1 =0 a= sec theta-tan theta b=cosec theta + cot theta ​

Answer 1

Answer:

I'm writing theta as p to reduce error.

a= secp-tanp = (1-sinp)/cosp

b= cosecp+cotp = (1+cosp)/sinp

then, ab = (1-sinp)(1+cosp)/sinpcosp

= (1+cosp-sinp-sinpcosp)/sinpcosp

a-b = [sinp(1-sinp)-cosp(1+cosp)]/sinpcosp

= (sinp-sin^2p-cosp-cos^2p)/sinpcosp

= (sinp-cosp-(sin^2p+cos^2p)/sinpcosp

= (sinp-cosp-1)/sinpcosp

Now, LHS= ab+a-b+1

= [(1+cosp-sinp-sinpcosp)/sinpcosp]+[(sinp-cosp-1)/sinpcosp]+1

= [(1+cosp-sinp-sinpcosp+sinp-cosp-1)/sinpcosp]+1

= (-sinpcosp/sinpcosp)+1

= -1+1

= 0

= RHS