IIT-JEE 1993Question 1
Determine the smallest positive value of number $x$ (in degrees) for which $$\tan\left(x+100^\circ\right)=\tan\left(x+50^\circ\right)\tan\left(x\right)\tan\left(x-50^\circ\right).$$
Trigonometric Ratio & Identities
Answer$30^\circ$
Solution
Concept. Move $\tan(x-50^\circ)$ to the left. Both sides then become ratios in $S=\sin(2x+50^\circ)$ and $C=\cos(2x+50^\circ)$. Formulas used. (1) $2\sin P\cos Q=\sin(P+Q)+\sin(P-Q)$, $2\cos P\sin Q=\sin(P+Q)-\sin(P-Q)$(2) $2\sin P\sin Q=\cos(P-Q)-\cos(P+Q)$, $2\cos P\cos Q=\cos(P-Q)+\cos(P+Q)$Step 1 - Rearrange. $$\frac{\tan(x+100^\circ)}{\tan(x-50^\circ)}=\tan(x+50^\circ)\tan x.$$ Step 2 - Left side. By (1), with $(x+100^\circ)-(x-50^\circ)=150^\circ$ and $\sin150^\circ=\frac12$: $$\frac{\sin(x+100^\circ)\cos(x-50^\circ)}{\cos(x+100^\circ)\sin(x-50^\circ)}=\frac{S+\frac12}{S-\frac12}.$$ Step 3 - Right side. By (2): $$\frac{\sin(x+50^\circ)\sin x}{\cos(x+50^\circ)\cos x}=\frac{\cos50^\circ-C}{\cos50^\circ+C}.$$ Step 4 - Cross-multiply. The terms $S\cos50^\circ$ and $\frac C2$ cancel, leaving $$2SC+\cos50^\circ=0\ \Longrightarrow\ \sin(4x+100^\circ)=-\sin40^\circ=\sin220^\circ.$$ Step 5 - Smallest positive $x$. $4x+100^\circ=220^\circ+360^\circ k$ gives $x=30^\circ+90^\circ k$; $4x+100^\circ=-40^\circ+360^\circ k$ gives $x=-35^\circ+90^\circ k$. The smallest positive value is $x=30^\circ$ (all tangents are defined there). Answer: $30^\circ$