Question

write the value of (1+tan2theta) (1+sin theta) (1_ sin theta) .

Answer 1

(1+tan2)(1+sin)(1-sin)=(sec2)(1-sin2)
=(sec2)(cos2)=(1/cos2)(cos2)=1

Answer 2

= (1+tan^2A)(1+sinA)(1-sinA)
= (1+tan^2A)(1-sin^2A)                          {(a+b)(a-b) = a^2-b^2}
= (sec^2A)(cos^2A)
= (1/cos^2A)(cos^2A)
= 1