Question

prove that sec0+tan0=cos0/1-sin0​

Answer 1

Step-by-step explanation:

LHS=sec0+tan0

=(1/cos0)+(sin0/cos0)

=(1+sin0)/cos0

=(1-sin^20)/{cos0(1-sin0)}

=cos^20/{cos0(1-sin0)}

=cos0/1-sin0

=RHS

Answer 2

MATHS ARYABHATA HERE

YOUR ANSWER:-

sec0+tan0

LHS=sec0+tan0 =(1/cos0)+(sin0/cos0)

LHS=sec0+tan0 =(1/cos0)+(sin0/cos0) =(1+sin0)/cos0

LHS=sec0+tan0 =(1/cos0)+(sin0/cos0) =(1+sin0)/cos0 =(1-sin^20)/{cos0(1-sin0)}

LHS=sec0+tan0 =(1/cos0)+(sin0/cos0) =(1+sin0)/cos0 =(1-sin^20)/{cos0(1-sin0)} =cos^20/{cos0(1-sin0)}

LHS=sec0+tan0 =(1/cos0)+(sin0/cos0) =(1+sin0)/cos0 =(1-sin^20)/{cos0(1-sin0)} =cos^20/{cos0(1-sin0)} =cos0/1-sin0

LHS=sec0+tan0 =(1/cos0)+(sin0/cos0) =(1+sin0)/cos0 =(1-sin^20)/{cos0(1-sin0)} =cos^20/{cos0(1-sin0)} =cos0/1-sin0 =RHS

#BE BRAINLY

SMART SOLVER.....