<s>[INST] Find $k$ if Sum of all Trigonometric ratios is $7$ Given $0< A< 90^{\circ}$ and $$\sin A+\cos A+\tan A+\sec A+\operatorname{cosec} A+\cot A=7$$
and if $\sin A$ and $\cos A$ are roots of $4x^2+3x+k=0$ 
Find the value of $k$
sum of the roots is $$\sin A+\cos A=\frac{-3}{4}$$ 
Squaring both sides we get
$$1+2\sin A\cos A=\frac{9}{16}$$ $\implies$
$$\sin A\cos A=\frac{-7}{32}$$
But Product of roots is
$$\sin A\cos A=\frac{k}{4}$$
so $$\frac{k}{4}=\frac{-7}{32}$$ Hence
$$k=\frac{-7}{8}$$
But it is also given that
$$\sin A+\cos A+\tan A+\sec A+\operatorname{cosec}A+\cot A=7$$ $\implies$
$$\sin A+\cos A+\frac{1}{\sin A\cos A}+\frac{\sin A+\cos A}{\sin A\cos A}=7$$ substituting sum of the roots and product of the roots we get
$$\frac{-3}{4}+\frac{4}{k}+\frac{-3}{k}=7$$ so
$$\frac{1}{k}=\frac{31}{4}$$ so
$$k=\frac{4}{31}$$
The two values are mismatching?
 [/INST] You're right. This equation is not right.
From this post: If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
it shows that $\sin x \cos x$ is a root of $x^2 -44x +36=0$
=> $\sin x \cos x$ = $22-8\sqrt{7}$ (Another value greater than 1 is ignored).
However, $(\sin x + \cos x)^2=1+2\sin x \cos x = 45-16 \sqrt{7} \neq 9/16$
Therefore there seems to be a problem in the equation given.
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