" (iii) "quad (tan1-i)^(2)

Find the modulus,argument,and the principal argument of the complex numbers.(i) (tan1-i)^(2)

" (i) "tan( 1/2(tan^(-1)x+tan^(-1)(1/x)))=1

Argument of the complex number (tan 1-i)^(2) is

Find the principal values of the following (i) tan^(-1) (tan. (2pi)/(3)) (ii) tan^(-1) (tan (-6))

Find the principal values of the following (i) tan^(-1) (tan. (2pi)/(3)) (ii) tan^(-1) (tan (-6))

" (i) "tan^(-1)1+tan^(-1)2+tan^(-1)3=pi

Pove that i) tan^(-1)1/2+tan^(-1)2/11=tan^(-1)3/4 ii) tan^(-1)2/11+tan^(-1)7/24=tan^(-1)1/2 iii) tan^(-1)1+tan^(-1)1/2+tan^(-1)1/3=pi/2 iv) 2tan^(-1)1/3+tan^(-1)/17=pi/4 v) tan^(-1)2-tan^(-1)1=tan^(-1)1/3 vi) tan^(-1)+tan^(-1)2+tan^(-1)3=pi vii) tan^(-1)1/2+tan^(-1)1/5+tan^(-1)1/8=pi/4 viii) tan^(-1)1/4+tan^(-1)2/9=1/2tan^(-1)4/3

The solution of the inequality (tan^(-1)x)^(2)-3tan^(-1)x+2>=0 is- a.(-oo,tan1]uu[tan2,oo)b*(-oo,tan1] c.(-oo,-tan1]uu[tan2,oo)d.[-1,1]-(-(sqrt(2))/(2),(sqrt(2))/(2))

Prove that (i) "cos " 15^(@) - " sin " 15^(@) = (1)/(sqrt(2)) (ii) " cot " 105^(@) - " tan " 105^(@) =2sqrt(3) (iii) (tan 69^(@) + tan 66^(@))/(1-tan 69^(@) tan 66^(@)) =-1

If 2A is not an integral multiple of pi , then show that i) cot A+tan A=2"cosec " 2A ii) cot A-tan A=2 cot2A and deduce the values of tan52 1^(@)/2 and tan37 1^(@)/2