Asked by Roy
                1) Simplification (trigonometric ratios of allied angles)?
Simplify :
i) [cos^2 (2x180-A). tan^2 (180+A). sec^2 (180-A)] / [ sin^2 (3x180+ A). cosec^2 (-A). cot^2 (90+A)]
ii)[ cos(90+A). sec(270+A). sin(180-A) ] / [ cosec (-A). cos (270-A). tan (180+A) ]
Find values of x and y lying between 0 degree and 360 degree?
2) find values of x and y lying between 0 degree and 360 degree
i) sin 2x = 0. 6428
ii) cos (1/2y+71 degree) = -0 . 3420
            
        Simplify :
i) [cos^2 (2x180-A). tan^2 (180+A). sec^2 (180-A)] / [ sin^2 (3x180+ A). cosec^2 (-A). cot^2 (90+A)]
ii)[ cos(90+A). sec(270+A). sin(180-A) ] / [ cosec (-A). cos (270-A). tan (180+A) ]
Find values of x and y lying between 0 degree and 360 degree?
2) find values of x and y lying between 0 degree and 360 degree
i) sin 2x = 0. 6428
ii) cos (1/2y+71 degree) = -0 . 3420
Answers
                    Answered by
            Steve
            
    cos^2(360-A) = cos^2(A)
tan^2(180+A) = tan^2(A)
sec^2(180-A) = sec^2(A)
sin^2(540+A) = sin^2(A)
csc^2(-A) = csc^2(A)
cot^2(90+A) = tan^2(A)
so, using those simplifications, we have
cos^2 * tan^2 * sec^2 = tan^2
sin^2 * csc^2 * tan^2 = tan^2
and the fraction is just 1
How far do you get with the others?
    
tan^2(180+A) = tan^2(A)
sec^2(180-A) = sec^2(A)
sin^2(540+A) = sin^2(A)
csc^2(-A) = csc^2(A)
cot^2(90+A) = tan^2(A)
so, using those simplifications, we have
cos^2 * tan^2 * sec^2 = tan^2
sin^2 * csc^2 * tan^2 = tan^2
and the fraction is just 1
How far do you get with the others?
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