Q.Find the values of other five trigonometric functions if $\sin x = \frac{3}{5}$, $x$ lies in second quadrant.

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): $\tan 45^\circ \overset{?}{=} \tan(-315^\circ)$.

Tangent has period $180^\circ$, and also $360^\circ$ works. Add one full revolution to $-315^\circ$:

So both sides equal $1$ ✓ — TRUE.

Step 5 — Answer the question actually asked.

Three statements are correct; the one that is not correct is (A). Read the stem carefully — it asks for the false statement, so the answer is the odd one out.

ANSWER: A

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.

Step 4 — Geometric cross-check. For $\tan\theta = 2$, take a right triangle with opposite $=2$, adjacent $=1$, hypotenuse $=\sqrt5$; then $\sec\theta = \sqrt5/1$ and $\sec^2\theta = 5$ ✓. For $\cot\phi = 3$, take adjacent $=3$, opposite $=1$, hypotenuse $=\sqrt{10}$; then $\csc\phi = \sqrt{10}/1$ and $\csc^2\phi = 10$ ✓.

ANSWER: C

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:

x sin A = sec A - cos A = (1 - cos^2 A)/cos A = sin^2 A / cos A

x = sin A / cos A = tan A.

(Valid wherever sin A is non-zero, which is required for the equation to determine x.)

ANSWER: C

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.

Moving left→right across the quadrant, $\tan x$ climbs from $-\infty$ up to $0$ — always negative, always rising. Sample values confirm it:

Each is larger than the last. ✓

Step 5 — Eliminate.

• (A) 'decreases from $-\infty$ to 0' — the values are right but the derivative says increases. ✗
• (B) 'increases from 0 to $\infty$' — that is the first-quadrant behaviour. ✗
• (D) 'decreases from 0 to $\infty$' — self-contradictory. ✗
• (C) increases from $-\infty$ to 0 ✓

ANSWER: C

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:

5. Take the limit by substituting $x=\pi$, i.e. $\cos\pi = -1$:

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.

Alternative (identity route, no root-finding).

since $\tan A\cot A=1$. Then

ANSWER: A