cos^(4)A-sin^(4)7i=

Period of 6cos^(4)x-7sin^(4)x is

Period of 6cos^(4)(x)-7sin^(4)(x) is

"(i) "sin^(4)theta-cos^(4)theta=sin^(2)theta-cos^(2)theta

sin4A=4cos^(3)A sin A-4sin^(3)A cos A

(i) int 1/ (cos^4x+sin^4x)dx (ii) int 1/(sin^4x+sin^2x cos^2x+cos^4x) dx

If (cos^(4)A)/(cos^(2)B)+(sin^(4)A)/(sin^(2)B)=1 then prove that (i)sin^(2)A+sin^(2)B=2sin^(2)A sin^(2)B(ii)(cos^(4)B)/(cos^(2)A)+(sin^(4)B)/(sin^(2)A)=1

Solve the following trigonometric equations, (i) 4 cos^(2)x+6"sin"^(2)x=5 (ii) 7 "sin"^(2)x+3 cos^(2)x=4 .

Using application of trignometric formulas prove that (i)cos(pi/4+x)+cos(pi/4-x)=sqrt2cos x(i1)sin(7pi/12)cos(pi/4)-cos(7pi/12)sin(pi/4)

sin((7pi)/(12))cos(pi/4)-cos((7pi)/(12))sin(pi/4)=...........