if tan a/2= square root (1+k/1-k)tan b/2 then sec a (cos b - k)/1-kcos b

tana/2=√(1-k/(1+k) tanb/2
or, tanb/2=√(1+k)/(1-k) tana/2
squaring both sides, 
tan²b/2=(1+k)/(1-k) tan²a/2
or, (1-tan²b/2)/(1+tan²b/2)={(1-k)-(1+k)tan²a/2}/{(1-k)+(1+k)tan²a/2}
[by dividendo-componendo method]
or, cosb=(1-k-tan²a/2-ktan²a/2)/(1-k+tan²a/2+ktan²a/2)
or, cosb{(1-tan²a/2)-k(1+tan²a/2)}/{(1+tan²a/2)-k(1-tan²a/2)}
or, cosb=[{(1-tan²a/2)-k(1+tan²a/2)}/(1+tan²a/2)]/
                                                   [{(1+tan²a/2)-k(1-tan²a/2)}/(1+tan²a/2)]
or, cosb=(cosa-k)/(1-kcosa) [∵, (1-tan²a/2)/(1+tan²a/2)=cosa] 

 

 

hence on side transwpfer you will get amswer as 1