Find $k$ if Sum of all Trigonometric ratios is $7$

Question

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?

I think I have even worse news for you: if $ \ A \ $ is in the first quadrant, how would $ \ \sin A \ + \ \cos A \ $ be negative? (Are you sure you have the conditions posted correctly?) — colormegone, Jun 07 '16 at 05:00

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

You're right. This equation is not right.

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.

1 Answers1