Q.Find the value of the trigonometric function $\csc(-1410^\circ)$.

The concept. Sine and cosine have period $2\pi$; tangent has period $180^\circ$ ($=\pi$). An odd multiple of $\pi$ lands cosine at $-1$, an even multiple at $+1$. Sine is zero at every integer multiple of $\pi$. Test each option; the one that fails is the answer.

Step 1 — Option (A): $\cos 5\pi \overset{?}{=} \cos 4\pi$.

$5\pi$ is an odd multiple of $\pi$, so

$4\pi$ is an even multiple of $\pi$ (a full two revolutions), so

Since $-1 \neq +1$, this statement is FALSE. This is our candidate.

Step 2 — Option (B): $\sin 2\pi \overset{?}{=} \sin(-2\pi)$.

$0 = 0$ ✓ — TRUE. (Sine is odd, but the negative of zero is still zero.)

Step 3 — Option (C): $\sin 4\pi \overset{?}{=} \sin 6\pi$.

Sine vanishes at every integer multiple of $\pi$:

✓ — TRUE.

Step 4 — Option (D): $\tan 45^\circ \overset{?}{=} \tan(-315^\circ)$. …

Step 1 — First term. Let $\theta = \tan^{-1}2$, so $\tan\theta = 2$ (and $\theta \in (0, \pi/2)$, the principal branch). Using the identity

Step 2 — Second term. Let $\phi = \cot^{-1}3$, so $\cot\phi = 3$ (with $\phi \in (0, \pi/2)$). Using

Step 3 — Add.

$5 + 10 = 15.$ …

We are given $\sin A = \frac{4}{5}$ with $A$ in the first quadrant, and $\cos B = -\frac{12}{13}$ with $B$ in the third quadrant. The goal is $\cos(A+B)$.

Concept & Intuition

The formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$ is straightforward — the catch is that we don't yet know $\cos A$ or $\sin B$. Their signs are fixed by the quadrants; compute their magnitudes with the Pythagorean identity and attach the correct sign.

Solution

1. Find $\cos A$. $A$ in QI $\Rightarrow \cos A > 0$.

2. Find $\sin B$. $B$ in QIII $\Rightarrow \sin B < 0$.

3. Apply the formula. …

Concept: allied-angle (co-function) identities for a 90 deg shift.

Step 1 - reduce each allied ratio:

• cosec(90 + A) = sec A
• cot(90 + A) = -tan A
• sin(90 + A) = cos A

Step 2 - substitute into the given relation:

sec A + x cos A (-tan A) = cos A

sec A - x cos A * (sin A / cos A) = cos A

sec A - x sin A = cos A

Step 3 - solve for x: …

Step 1 — Fix the interval.

The second quadrant means

Step 2 — Establish the direction (increasing or decreasing) rigorously.

Differentiate:

A square is never negative, and $\cos x \ne 0$ on the open interval, so

A positive derivative ⇒ the function is strictly increasing. This immediately kills options (A) and (D), which both say 'decreases'. (Note: $\tan x$ is increasing on every one of its branches — it never decreases anywhere.)

Step 3 — Find the two end-values (the range on this branch).

Use $\tan x = \dfrac{\sin x}{\cos x}$ and the signs in Q-II: $\sin x > 0$, $\cos x < 0$, so $\tan x < 0$ throughout the quadrant.

• As $x \to \left(\tfrac{\pi}{2}\right)^{+}$: $\sin x \to 1^{-}$ (positive) while $\cos x \to 0^{-}$ (a small negative number). A positive divided by a vanishing negative gives

• As $x \to \pi^{-}$: $\sin x \to 0^{+}$ and $\cos x \to -1$, so

Step 4 — Combine. …

The matrix $f(x)$ depends on $x$ only through $\cos x$, so its determinant $|f(x)|$ is a polynomial in $\cos x$ — and therefore continuous. Because it is continuous, the limit as $x\to\pi$ is simply its value at $x=\pi$, where $\cos\pi=-1$.

Let's work through the determinant step by step.

1. Write the matrix:

2. Expand along the first column. Its entries are $\cos x$, $0$, $0$, so only the first term survives:

3. Compute the $2\times2$ determinant:

4. Therefore: …

Step 1 — Reduce to one variable.

Since $\cot A=\dfrac{1}{\tan A}$, set $t=\tan A$ ($t\neq0$). The condition becomes

Step 2 — Solve.

Multiply through by $t$:

So $\tan A=1$ (e.g. $A=45^\circ$) and consequently $\cot A=\dfrac{1}{1}=1$.

(The underlying idea is the AM–GM equality case: for $t>0$, $t+\frac1t\ge2$, with equality only at $t=1$. The given equation sits exactly at that boundary, which is why the quadratic has a repeated root.)

Step 3 — Evaluate the required expression. …