Basics & LogarithmshardFree

Telescoping Sum of Logarithms of Tangents from 1 to 89 Degrees | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
If ax=ba^{x} = b, by=cb^{y} = c, cz=ac^{z} = a (where a,b,c,x,y,za, b, c, x, y, z are positive real numbers), then the value of
∑r=189log⁡10tan⁡(rπ180)xyz\frac{\displaystyle\sum_{r=1}^{89}\log_{10}\tan\left(\frac{r\pi}{180}\right)}{xyz}
 is equal to
Solution
Answer: 0
Step 1: Chain the three relations to find xyzxyz. Substituting b=axb = a^{x} into by=cb^{y} = c,
c=(ax)y=axy,c = \left(a^{x}\right)^{y} = a^{xy} ,
 and then substituting into cz=ac^{z} = a,
(axy)z=a ⟹ axyz=a1.\left(a^{xy}\right)^{z} = a \ \Longrightarrow\ a^{xyz} = a^{1} .
 Step 2: Conclude. Since aa is a positive real other than 11 (otherwise the relations are vacuous), the exponents must match:
xyz=1.xyz = 1 .
 So the denominator is 11 - and, importantly, it is not zero, so the quotient is defined.Step 3: Turn to the numerator, which is a sum of logarithms of tangents of 1∘1^{\circ} through 89∘89^{\circ}. Recall rπ180\dfrac{r\pi}{180} is simply rr degrees. Step 4: Pair the terms from the two ends, using tan⁡(90∘−θ)=cot⁡θ\tan(90^{\circ}-\theta) = \cot\theta.
tan⁡r∘⋅tan⁡(90−r)∘=tan⁡r∘⋅cot⁡r∘=1,\tan r^{\circ}\cdot\tan(90-r)^{\circ} = \tan r^{\circ}\cdot\cot r^{\circ} = 1 ,
 so
log⁡10tan⁡r∘+log⁡10tan⁡(90−r)∘=log⁡101=0.\log_{10}\tan r^{\circ} + \log_{10}\tan(90-r)^{\circ} = \log_{10}1 = 0 .
 Step 5: Count what the pairing covers. The values r=1r = 1 to 8989 pair up as (1,89),(2,88),…,(44,46)(1,89), (2,88), \ldots, (44,46) - that is 4444 pairs, each summing to zero - leaving only r=45r = 45 unpaired. Step 6: Handle the leftover term.
log⁡10tan⁡45∘=log⁡101=0.\log_{10}\tan45^{\circ} = \log_{10}1 = 0 .
 Step 7: Total.
numerator=0,01=0.\text{numerator} = 0 , \qquad \frac{0}{1} = 0 .
 Answer: 00.
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
If x=∏n=12026nx = \prod_{n=1}^{2026} n, then the value of the expression 11log⁡2x+1log⁡3x+⋯+1log⁡2026x\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x} + \cdots + \dfrac{1}{\log_{2026} x}} is
Basics & Logarithms · medium
Sum of all possible values of xx of the equation (3+2)x+(3−2)x−23=0(\sqrt{3}+\sqrt{2})^x + (\sqrt{3}-\sqrt{2})^x - 2\sqrt{3} = 0 is:
Basics & Logarithms · easy
Number of integral values of xx satisfying the inequality (34)6x+10−x2<2764\left(\dfrac{3}{4}\right)^{6x+10-x^2} < \dfrac{27}{64} is:
Basics & Logarithms · medium
Find the number of negative integers satisfying −1≤x2−5x+4x2−4≤1-1 \leq \dfrac{x^2-5x+4}{x^2-4} \leq 1 is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath