Q.Integrate the following function: $\cos 2x \cos 4x \cos 6x$

Step 1 — Rearrange the given condition.

We are given

Isolate $\cos x$:

Step 2 — The concept: use the Pythagorean identity.

The fundamental identity $\sin^2 x + \cos^2 x = 1$ rearranges to

The right-hand side of Step 1 is exactly this. So the given condition is equivalent to the elegant relation

This is the whole trick — the condition secretly says "$\cos x$ is $\sin^2 x$", which lets us swap one for the other.

Step 3 — Rewrite the required expression in terms of $\sin^2 x$.

Step 4 — Substitute $\sin^2 x = \cos x$.

Step 5 — Recognise the given condition and finish.

But $\cos x + \cos^2 x = 1$ is precisely what we were told. Therefore

Step 6 — Numerical sanity check.

Solve $\cos^2 x + \cos x - 1 = 0$ for $c=\cos x$: $c = \dfrac{-1+\sqrt{5}}{2} \approx 0.6180$ (the root in $[-1,1]$).

Then $\sin^2 x = 1 - c^2 \approx 1 - 0.3820 = 0.6180$ (indeed $=c$ ✓), and $\sin^4 x \approx 0.3820$.

Sum: $0.6180 + 0.3820 = 1.0000$ ✓ — matches.

Also note the value must be positive (a sum of even powers), which immediately kills (A) $-1$; and it cannot be $0$ (that would need $\sin x = 0$, forcing $\cos x = \pm1$, neither of which satisfies the condition), killing (C).

ANSWER: B

The core idea here is that the given limit looks exactly like the definition of a derivative, but with a symmetric difference. Instead of blindly applying L'Hôpital's rule, recognise that

is a standard form for $2f'(2)$, provided $f$ is differentiable. Here $f(t) = \sin t$, so $f'(t) = \cos t$. That gives the answer almost instantly.

Let’s walk through it carefully.

1. Identify the function.

   Let $f(t) = \sin t$. Then the numerator is $f(2+x) - f(2-x)$.

2. Rewrite the limit in derivative form.

   The derivative of $f$ at $t=2$ is

But our limit uses $f(2-x)$ instead of $f(2)$. Notice that

Split it:

The second term can be rewritten by letting $h = -x$:

where $h = -x$. As $x \to 0$, $h \to 0$ as well. So both terms approach $f'(2)$.

3. Combine the two pieces. Hence

4. Compute $f'(2)$.

   Since $f(t) = \sin t$, $f'(t) = \cos t$, so $f'(2) = \cos 2$.

5. Match to the given form.

   The limit equals $2 \cos 2$. The problem says this equals $A \cos B$. So $A = 2$ and $B = 2$.

The key insight is that in a right triangle, the two acute angles add up to $90^\circ$. That means $A + B = 90^\circ$, so $B = 90^\circ - A$. This immediately tells us $\tan B = \cot A$, because $\tan(90^\circ - A) = \cot A$. So the sum $\tan A + \tan B$ becomes $\tan A + \cot A$.

But the options are given in terms of side lengths $a$, $b$, $c$ — where by standard convention, side $a$ is opposite $A$, side $b$ opposite $B$, and side $c$ opposite $C$ (the hypotenuse, since $C = 90^\circ$). So we need to express $\tan A + \cot A$ in terms of these sides.

Let’s work through it step by step.

1. Set up the triangle.

   Right angle at $C$ means $c$ is the hypotenuse. So $AB = c$, $BC = a$ (opposite $A$), and $AC = b$ (opposite $B$).

   From the definitions:

   $\tan A = \frac{\text{opposite to }A}{\text{adjacent to }A} = \frac{a}{b}$

   $\tan B = \frac{\text{opposite to }B}{\text{adjacent to }B} = \frac{b}{a}$

2. Add them directly.

   $\tan A + \tan B = \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$

3. Use the Pythagorean theorem.

   Since the triangle is right-angled at $C$, we have $a^2 + b^2 = c^2$.

   Therefore $\tan A + \tan B = \frac{c^2}{ab}$.

That matches option (C).