Determine the sign of cos pi divided by seven. and Is the function cot t positive or negative in Quadrant II?

1) pi/7 is in quad I and all trig functions are positive there, so positive 

2) cot is negative in II

cos^4a+cos^2a=1 ,prove is it identity or not

not an identity:
To disprove an identity we need only a single exception: 
 
and not 1 

:)

sec^4 a+sec^2 a=1 ,prove is it identity or not,thanks and regards

clearly not an identity try a=pi, substitution leads to 
2=1 
:)

(a+5b-4)^3

(a+5b-4)^3= 


(a+5b-4)(a+5b-4)(a+5b-4)= 


(a+5b-4)[a(a+5b-4)+5b(a+5b-4)-4(a+5b-4)]= 

(a+5b-4)[a^2+5ab-4a+5ab+25b^2-20b-4a-20b+16]= 

(a+5b-4)[a^2+10ab+25b^2-8a-40b+16]= 
a[a^2+10ab+25b^2-8a-40b+16]+
5b[a^2+10ab+25b^2-8a-40b+16]-
4[a^2+10ab+25b^2-8a-40b+16]= 


a^3+15a^2b-12a^2+75ab^2-120ab+48a+125b^3-300b^2+240b-64

solve 7x^2+10x=8

7x^2+10x=8
7x^2+10x-8=0
7x^2+14x-4x-8=0
7x(x+2)-4(x+2)=0
(7x-4)(x+2)=0 
x=-2 or x=4/7 

:)

A 6 character computer password is made up of 4 numbers followed by 2 letters. How many different passwords are possible?

counting principle applies: 
(10)(10)(10)(10)(26)(26)=6760000 

:)

Can you help me solve for real roots? 2x^8+15x^4=27

u=x^4 
2u^2+15u-27=0 
2u^2+18u-3u-27=0 
2u(u+9)-3(u+9)=0 
(2u-3)(u+9)=0 

but u+9 contributes no real roots 

2u-3=0 
u=3/2 

x^4 = 3/2 

x^2 = +- sqrt(3/2) 

from which we use the principal root 

thus 
 
or