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    (Original post by shushmush)
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    Have you done part (i)?
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    Yes
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    (Original post by shushmush)
    Yes
    You know that \tan^2 {\theta}-\sin^2{\theta}>0 since \tan{\theta}>0 and \sin{\theta}>0 for 0^\circ<\theta<90^\circ. Try factorising the expression and tell me what you think.
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    (Original post by shushmush)
    Yes
    Sorry, didn't see this as I wasn't quoted.

    So because the interval is 0 < x < 90, you know that the RHS is always going to be positive as both sin(x) and tan(x) are both positive for this range, so sin2(x) and tan2(x) will be too.

    Factorise the LHS and you'll probably find your answer
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    (Original post by Protostar)
    Sorry, didn't see this as I wasn't quoted.

    So because the interval is 0 < x < 90, you know that the RHS is always going to be positive as both sin(x) and tan(x) are both positive for this range, so sin2(x) and tan2(x) will be too.

    Factorise the LHS and you'll probably find your answer
    I'm not sure how to factorise the LHS.
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    (Original post by shushmush)
    I'm not sure how to factorise the LHS.
    Are you familiar with the difference of two squares?
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    Ohhhh I get the factorisation now
    (Original post by Protostar)
    Are you familiar with the difference of two squares?
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    (Original post by shushmush)
    Ohhhh I get the factorisation now
    Great so what do you spot?
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    (Original post by Protostar)
    Great so what do you spot?
    What am I meant to spot?
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    (Original post by shushmush)
    What am I meant to spot?
    We've established that the RHS is always positive for values of x in this range. Because of the identity, the LHS must also be positive. Look at the result of your factorisation, keeping this in mind, and things should become reasonably clear
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    (Original post by Protostar)
    We've established that the RHS is always positive for values of x in this range. Because of the identity, the LHS must also be positive. Look at the result of your factorisation, keeping this in mind, and things should become reasonably clear
    Thank you!
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    (Original post by shushmush)
    Thank you!
    No worries, hope that helped
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    tan x is sinx/cosx

    cosx is between 0 and 1 not inclusive

    if you divide by a positive number less than 1 your answer gets bigger.
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    (Original post by shushmush)
    ...
    You've established that (\tan(x)+\sin(x))(\tan(x)-\sin(x)) \equiv (\tan(x) \sin(x))^2

    We know that the RHS is  &gt; 0 for 0&lt; x &lt; 90 because it's a whole square term (anything squared is always +ve)

    So, you essentially have (\tan(x)+\sin(x))(\tan(x)-\sin(x)) &gt; 0 for 0&lt; x &lt; 90

    We can simplify this by dividing by one of the expressions, but you need to reason WHY you're allowed to divide by one of them. Which would leave you with the other expression and the inequality, thus granting the result.
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    (Original post by RDKGames)
    We know that the RHS is  &gt; 0 for 0&lt; x &lt; 90 because it's a whole square term (anything squared is always +ve)
    Just a pedantic quibble on language but real numbers squared are always non-negative, not necessarily positive; you can get zero, and the fact that you don't in this case is important to the argument.
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    (Original post by I hate maths)
    Just a pedantic quibble on language but anything squared is always non-negative, not necessarily positive; it can be zero, and the fact that it's not in this case is important to the argument.
    For more pedantry: any real number squared is always non-negative; i^2 = -1.
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    (Original post by I hate maths)
    Just a pedantic quibble on language but anything squared is always non-negative, not necessarily positive; it can be zero, and the fact that it's not in this case is important to the argument.
    Yes anything (real) squared is non-negative, but in this context we have 0&lt;x&lt;90, so the product squared is strictly positive as I said.
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    (Original post by Zacken)
    For more pedantry: any real number squared is always non-negative; i^2 = -1.
    Ah yeah, edited it immediately after it went up but it seems it wasn't on time.
 
 
 
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