NDA 2026 — Mathematics PYQ

To solve this, we use the change-of-base property for logarithms: ${\log }_{a}b=\frac{1}{{\log }_{b}a}$.

Step 1: Simplify the equation

Let $y={\log }_{\sin x}\cos x$. Then the given equation becomes:

$$y+\frac{1}{y}=2$$

Step 2: Solve for $y$

Multiply by $y$:

$$y^2-2y+1=0$$

$$(y-1{)}^2=0⟹y=1$$

Step 3: Relate back to trigonometric functions

Since $y={\log }_{\sin x}\cos x=1$, by the definition of logarithms:

$$\cos x=(\sin x{)}^1$$

$$\cos x=\sin x$$

Step 4: Solve for $x$

Divide both sides by $\cos x$ (assuming $\cos x\ne 0$):

$$\tan x=1$$

For the smallest positive value of $x$:

$$x={\tan }^{-1}(1)=\frac{\pi }{4}$$

Final Answer:

The smallest positive value is $x=\frac{\pi }{4}$. The correct option is (c).