IPMAT Indore 2023 · Quantitative Ability · Question 36

Step 1 — use the reciprocal relation between the two logs.

{\log}_{\sin x}\left( \cos x \right)=\frac{1}{{\log}_{\cos x}\left( \sin x \right)}

So writing t={\log}_{\cos x}\left( \sin x \right), the equation becomes

t+\frac{1}{t}=2

Step 2 — solve for t.

{t}^{2}-2t+1=0 \Rightarrow{\left( t-1 \right)}^{2}=0 \Rightarrow t=1

Step 3 — unwind.

{\log}_{\cos x}\left( \sin x \right)=1 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1

The general solution of \tan x=1 is x=n\pi +\pi /4.

Step 4 — apply the domain restrictions. This is the step that decides the answer. For the logarithms to exist we need

\sin x>0, \cos x>0, \sin x\ne 1, \cos x\ne 1

(a logarithm needs a positive argument, and a base that is positive and not 1).

So both \sin x and \cos x must be positive, confining x to the first quadrant.

Of the solutions x=n\pi +\pi /4, the odd ones (x=5\pi /4, etc.) lie in the third quadrant where both are negative — invalid. Only the first-quadrant family survives:

x=2n\pi +\frac{\pi}{4}

Answer: Option 2 — 2n\pi +\frac{\pi}{4}

Shortcut. Two moves, and the second is the question.

t+1/t=2 forces t=1, so \tan x=1.

Then check the domain. A log needs a positive argument and a positive base \ne 1, so \sin x>0 and \cos x>0 — first quadrant only.

\tan x=1 gives n\pi +\pi /4, but half of those sit in the third quadrant. Discarding them leaves 2n\pi +\pi /4.

Option (c) is the trap, and it is exactly what you get by solving the trigonometry and skipping the domain. When the variable sits in a logarithm’s base, the domain check is not a formality — it is usually the whole point.