Question

prove that
{(1-sec∆+tan∆)/(1+sec∆-tan∆)}={(sec∆+tan∆-1)/(sec∆+tan∆+1)}​

Answer 1

Answer:

Step-by-step explanation:

LHS=1-(secα-tanα)/1+(secα-tanα)  

= (sec²α-tan²α)-(secα-tanα)/(sec²α-tan²α)+(secα-tanα)

= ((secα+tanα)(secα-tanα) )-(secα-tanα) / ((secα+tanα)(secα-tanα) ) + (secα-tanα)

=[(secα-tanα)] [(secα+tanα)-1] / [(secα-tanα)] [(secα+tanα)+1]

=[ (secα+tanα)-1 ] / [(secα+tanα)+1] = RHS