[gozilla725]

have a take home quiz that I need to do well on. Having trouble with these:

Simplify. [1/(1+sinx)] + [1/(1-sinx)]

Factor and then simplify. (sec^2x-1)/(1+secx)

Factor then simplify. Cos^4x-2cos^2+1

Verify. (cotx tanx)/sinx=cscx

Verify. Sin^4x-cos^4x=2sin^2x-1


Really appreciate it.

[snowflake317]

Q.1 [1/(1+sinx)] + [1/(1-sinx)]

taking the LCM

= 1-sinx + 1 + sinx/ 1-sin^2x
= 2/cos^2x { since according to identity sin^2x +
cos^2x=1
=> 1- sin^2x = cos^2x}
= 2sec^2x { 1/cos^2x= sec^2x}



Q.2(sec^2x -1)/ (1+secx)

sec^2x -1 can be split into (secx-1)*(secx+1) { (a^2-b^2)= (a-b)*(a+b)}

= (secx-1)*(secx+1)/(1+secx)

= secx-1


Q.3 cos^4x-2cos^2x +1
= (cos^2x)^2 -(2cos^2x-1)
since 2cos^2x-1 = cos2x
& cos^2x= 1= sin^2x

= (1-sin^2x)^2 -( cos2x)
since cos2x= 1- 2sin^x

= 1 + sin^4x -2sin^x -(1 - 2sin^2x)
= 1+ sin^4x -2sin^x -1 + 2sin^2x
= sin^4x


Q.4 (cotx tanx)/sinx= cscx

taking the left side

since cotx= 1/tanx

= (tanx* 1/tanx)/ sinx
= 1/sinx
= cscx { since 1/sinx= cscx}

which equals the right side of the equation
hence, proved



Q.5 sin^4x - cos^4x= 2sin^2x-1

taking the left side

=sin^4x -(cos^2x)^2
=sin^4x -(1-sin^2x)^2
=sin^4x -(1 + sin^4x - 2sin^2x)
= sin^4x-1-sin^4x + 2sin^2x
= 2sin^2x-1

which equals the right side of the equation

hence, proved


I'm happy I could help you :)