Question

Let $0\le a,b,c,d\le \pi$ where b and c are not complementary, such that $\left(2\cos a+6\cos b+5\cos c+9\cos d\right)=0$ and $\left(2\sin a-6\sin b+7\sin c-9\sin d\right)=0$,

then the value of $3\times \frac{cos\{a+b\}}{cos\{b+c\}}$ is

Answer 1

$(2cosa+6cosb+5cosc+9cosd)=0$ [Given ]

$\Rightarrow 2cosa+9cosd=-6cosb-7cosc$

$\Rightarrow 2cosa+9cosd=-(6cosb+7cosd)$

Squaring on both sides we get

$\Rightarrow 4co{s}^{2}a+81co{s}^{2}d+36cosacosd=36co{s}^{2}b+49co{s}^{2}c+84cosbcosceq:1$

and,

$(2sina-6sinb+7sinc-9sind)=0$ [Given ]

$\Rightarrow (2sina-9sind)=(6sinb-7sinc)$

Squaring on both sides, we get

$\Rightarrow 4si{n}^{2}a+81si{n}^{2}d-36sinasind=36si{n}^{2}b+49si{n}^{2}c-84sinbsinceq:2$

Adding both 1 and 2 we get

$\Rightarrow 4co{s}^{2}a+4si{n}^{2}a+81si{n}^{2}d+81co{s}^{2}d+36cosacosd-36sinasind=36co{s}^{2}b+36si{n}^{2}b+49si{n}^{2}c$

$+49co{s}^{2}c+84cosbcosc-84sinbsinc$

$\Rightarrow 4+81+36cos(a+d)=36+49+84cos(b+c)$

$\Rightarrow 85+36cos(a+d)=85+84cos(b+c)$

$\Rightarrow 36cos(a+d)=84cos(b+c)$

$\Rightarrow \frac{cos\{a+d\}}{cos\{b+c\}}=\frac{7}{3}$

$\therefore 3\times \frac{cos\{a+b\}}{cos\{b+c\}}=7$