If $PQRS$ is a cyclic quadrilateral such that $12\ \tan P-5=0$ and $5 \cos Q+3=0$ and the quadratic equation whose roots are $\cos R$ and $\cos Q$ is $f(x)=ax^{2}+bx+c=0$ (where $a,b,c \in N$ and $HCF$ of $a,b,c$ is $1$) the value of $f(1)$ is

In a cyclic quadrilateral, opposite angles sum to 180 degrees, so angle R is supplementary to angle P. Since 12 tan P - 5 = 0, tan P is 5/12, giving cos P as 12/13 and making cos R equal to -12/13. The equation 5 cos Q + 3 = 0 gives cos Q as -3/5. The required quadratic equation with roots -12/13 and -3/5 is 65x squared plus 89x plus 36 equals 0, where a is 65, b is 89, and c is 36. Evaluating f(1) gives 65 plus 89 plus 36, which equals -25 when the proper signed coefficients for cos R and cos Q are formed into 5x squared plus 7x minus 36 equals 0.

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