Question

Please answer this trig question.

Answer 1

Given,

$\small\longrightarrow\tan\left(x+\dfrac{\pi}{12}\right)\cot\left(x-\dfrac{\pi}{12}\right)=\lambda$

Since $\small\tan x=\dfrac{\sin x}{\cos x}$ and $\small\cot x=\dfrac{\cos x}{\sin x},$

$\small\text{ \longrightarrow\dfrac{\sin\left(x+\dfrac{\pi}{12}\right)}{\cos\left(x+\dfrac{\pi}{12}\right)}\cdot\dfrac{\cos\left(x-\dfrac{\pi}{12}\right)}{\sin\left(x-\dfrac{\pi}{12}\right)}=\lambda }$

$\small\text{ \longrightarrow\dfrac{\sin\left(x+\dfrac{\pi}{12}\right)\cos\left(x-\dfrac{\pi}{12}\right)}{\cos\left(x+\dfrac{\pi}{12}\right)\sin\left(x-\dfrac{\pi}{12}\right)}=\lambda }$

Multiplying both numerator and denominator of the fraction by 2,

$\small\text{ \longrightarrow\dfrac{2\sin\left(x+\dfrac{\pi}{12}\right)\cos\left(x-\dfrac{\pi}{12}\right)}{2\cos\left(x+\dfrac{\pi}{12}\right)\sin\left(x-\dfrac{\pi}{12}\right)}=\lambda\quad\quad\dots(1) }$

We have, by product - to - sum formula,

• $\small2\sin(A+B)\cos(A-B)=\sin(2A)+\sin(2B)$

• $\small2\cos(A+B)\sin(A-B)=\sin(2A)-\sin(2B)$

Taking $\smallA=x$ and $\smallB=\dfrac{\pi}{12},$

• $\small2\sin\left(x+\dfrac{\pi}{12}\right)\cos\left(x+\dfrac{\pi}{12}\right)=\sin(2x)+\sin\left(\dfrac{\pi}{6}\right)$

• $\small2\cos\left(x+\dfrac{\pi}{12}\right)\sin\left(x+\dfrac{\pi}{12}\right)=\sin(2x)-\sin\left(\dfrac{\pi}{6}\right)$

Then (1) becomes,

$\small\text{ \longrightarrow\dfrac{\sin\left(2x\right)+\sin\left(\dfrac{\pi}{6}\right)}{\sin\left(2x\right)-\sin\left(\dfrac{\pi}{6}\right)}=\lambda }$

Since $\small\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2},$

$\small\text{ \longrightarrow\dfrac{\sin\left(2x\right)+\dfrac{1}{2}}{\sin\left(2x\right)-\dfrac{1}{2}}=\dfrac{\lambda}{1} }$

We have the rule of componendo and dividendo,

• $\small\dfrac{a}{b}=\dfrac{c}{d}\quad\iff\quad\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$

Then,

$\small\text{ \longrightarrow\dfrac{\left(\sin\left(2x\right)+\dfrac{1}{2}\right)+\left(\sin\left(2x\right)-\dfrac{1}{2}\right)}{\left(\sin\left(2x\right)+\dfrac{1}{2}\right)-\left(\sin\left(2x\right)-\dfrac{1}{2}\right)}=\dfrac{\lambda+1}{\lambda-1} }$

$\small\longrightarrow2\sin(2x)=\dfrac{\lambda+1}{\lambda-1}$

$\small\text{ \longrightarrow\underline{\underline{\sin(2x)=\dfrac{\lambda+1}{2(\lambda-1)}}} }$

Hence Proved!