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    The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k,

    (i) tan (1/2 pie - x)

    (ii) sin x


    the answer is

    (i) 1/k

    (ii) k/square root (1+k square)

    can anyone show me the complete working for this question? Thx
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    Use the tangent rule for tan(a-b) and simplfy i)
    For part ii) refer to Mr. Pythagoras
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    do you mind show me in detail workings? cause i dont really understand sorry
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    (Original post by CookieGhoul)
    Use the tangent rule for tan(a-b) and simplfy i)
    For part ii) refer to Mr. Pythagoras
    but tan(pi/2) is undefined? :confused:
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    1. \tan (\frac \Pi2 - x) = \cot x

    How about now ?
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    i appreciate it! thx for that! i think of the same thing as in question part 1 but how about part 2? that's a lot more complicated.
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    1. Let   \theta=\frac \Pi2 - x 

Therefore \tan \theta = \frac {\sin \theta}{\cos \theta}

Then : \sin (\frac \Pi2 - x) = \sin \frac \Pi2 \cos x - \cos \frac \Pi2 \ sinx

= \cos x 

And : \cos (\frac \Pi2 - x) = \cos \frac \Pi2 \cos x + \sin \frac \Pi2 \ sinx

= \sin x



Therefore we get \tan (\frac \Pi2 - x) = \cot x = \frac1{\tan x} = \frac1{k}
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    2.   Since \tan x = k, 

Hypotenuse = \sqrt (k^2+1). 

\sin x = \frac k{k^2+1}


    Just assume that opposite to angle x is the length k and the adjacent is 1
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    these won't come in p1 exams
 
 
 
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