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    The graphs of y=cos^-1(x) and
    y=tan^-1(x), where x lies between 1 and -1 in each case.


    The graphs intersect at the point with
    coordinates (p,q).


    Write down, in terms of p and q,the
    coordinates of the corresponding point of



    intersection of the graphs of y=cosx
    and y=tanx, and hence show that



    cos^2(q)=sinq. This is the question I
    tried and couldnt do will appreciate if
    someone helps me out with this.
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    Here's my take on it:
    y=cos^-1(x) => x=cos(y) ,
    y=tan^-1(x) => x=tan(y)
    x=tan(y)=cos(y) where they intersect.
    trig identities: Sin(y)/cos(y)=tan(y) => sin(y)/cos(y)=cos(y) => sin(y)= cos^2(y)
    and where they intersect p is the y coordinate, so just put the p in place of the y.
    Dunno about this corresponding point mumbo jumbo, wouldn't it just be (q,p)?
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    Ok, so, you know cos^-1x=y and tan^-1x=y.

    => cosy=x (1) and tany=x (2)

    Sub (1) into (2):

    =>cosy=tany

    =>cosy=siny/cosy

    Does this help?
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    (Original post by TheSK00T3R)
    Here's my take on it:
    y=cos^-1(x) => x=cos(y) ,
    y=tan^-1(x) => x=tan(y)
    x=tan(y)=cos(y) where they intersect.
    trig identities: Sin(y)/cos(y)=tan(y) => sin(y)/cos(y)=cos(y) => sin(y)= cos^2(y)
    and where they intersect p is the y coordinate, so just put the p in place of the y.
    Dunno about this corresponding point mumbo jumbo, wouldn't it just be (q,p)?
    Just ignore the word "corresponding"
 
 
 
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