theta sec(sin)/(d^(2)+-(1)/(1+1^(2)+1))+(2)/(1+2^(2)+2^(4))+--+(49)/(1+99^(2)+99^(7))=?

( The sum )/(1+1^(2)+1^(4))+(2)/(1+2^(2)+2^(4))+(3)/(1+3^(2)+3^(4))+.........+(99)/(1+99^(2)+99^(4))

If sec theta - tan theta =(a+1)/(a-1), then cos theta = ........... A) (a^(2) +1)/(a ^(2)-1) B) (a^(2) -1)/(a ^(2)+1) C) (2a)/(a ^(2)+1) D) (2a)/(a ^(2)-1)

If sec theta+tan theta=x, then sec theta=(a)(x^(2)+1)/(x) (b) (x^(2)+1)/(2x)(c)(x^(2)-1)/(2x) (d) (x^(2)-1)/(x)

{(99-1) // 7^(2)} xx 2+ 8 = ?

[(1)/(1*2)+(1)/(2*3)+(1)/(3*4)+....+(1)/(99*100)] =?

cot theta=(2tan7(1)/(2^(@)))/(1-tan^(2)7(1)/(2^(@))) then sin3 theta=

sin^(4) theta + 2 cos^(2) theta (1 - (1)/( sec^(2) theta)) + cos ^(4) theta =

(1-sin^(2)theta)sec^(2)theta=1

(1)/(1+sin^(2)theta)+(1)/(1+cos^(2)theta)+(1)/(1+sec^(2)theta)+(1)/(1+csc^(2)theta)=2

Prove that , (1+sec 2 theta) (1+sec2^(2) theta)(1+ sec2^(3)theta)... ( 1+sec 2^(2) theta) = (tan 2^(n) theta)/tan theta