Find the most general value of $\theta$ satisfying the equation $\tan\theta = -1$ and $\cos\theta = \dfrac{1}{\sqrt{2}}$.

Find the most general value of θ satisfying the equation tanθ = -1 and cosθ = dfrac1√(2).

NCERT Exemplar · Q16

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Concept understanding — Trigonometric Functions in Quadrants

Trigonometric Functions in Quadrants

Imagine standing at the centre of a circle, facing east. If you turn by some angle, you end up pointing in a certain direction. That direction has both a horizontal component (east-west) and a vertical component (north-south). Trigonometric functions are just a way to describe those components — and whether they are positive or negative depends entirely on which quadrant you're facing.

The Four Quadrants

The coordinate plane is split into four quadrants, numbered anticlockwise starting from the top-right:

Quadrant I (0° to 90°): x > 0, y > 0
Quadrant II (90° to 180°): x < 0, y > 0
Quadrant III (180° to 270°): x < 0, y < 0
Quadrant IV (270° to 360°): x > 0, y < 0

Now, recall the definitions on the unit circle (radius = 1):

$\cos \theta$ = x-coordinate of the point on the circle
$\sin \theta$ = y-coordinate of that point
$\tan \theta = \frac{\sin \theta}{\cos \theta}$

So the sign of $\cos \theta$ follows the sign of x, and the sign of $\sin \theta$ follows the sign of y. That's all there is to it.

The Sign Pattern

Quadrant	$\sin \theta$	$\cos \theta$	$\tan \theta$
I (0–90)	+	+	+
II (90–180)	+	–	–
III (180–270)	–	–	+
IV (270–360)	–	+	–

Tip

The mnemonic "All Students Take Coffee" helps you remember which functions are positive in each quadrant, starting from QI and going anticlockwise: All (all positive), Sine (sin positive), Tan (tan positive), Cos (cos positive).

Why This Matters

Suppose you're solving $\sin \theta = \frac{1}{2}$ . The calculator gives you $\theta = 30^\circ$ , but that's only one solution. Because sine is positive in both QI and QII, there's a second angle: $180^\circ - 30^\circ = 150^\circ$ . If you forget the quadrant rule, you lose half the answers.

Similarly, if $\cos \theta = -\frac{\sqrt{3}}{2}$ , cosine is negative in QII and QIII. So the solutions are $150^\circ$ and $210^\circ$ (plus full rotations). …

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We need $\theta$ where both $\tan\theta = -1$ and $\cos\theta = \frac{1}{\sqrt{2}}$ hold simultaneously. By analyzing signs in each quadrant, only the fourth quadrant satisfies both conditions, giving $\theta = 2n\pi - \frac{\pi}{4}$ or equivalently $\theta = (8n-1)\frac{\pi}{4}$ where $n \in \mathbb{Z}$ .

The heart of this problem lies in understanding that trigonometric equations don't just ask "where does this function equal this value?" but rather "where do all these conditions hold at once?" Each equation individually has infinitely many solutions, but their intersection may be much smaller.

When $\tan\theta = -1$ , the tangent is negative. When $\cos\theta = \frac{1}{\sqrt{2}}$ , the cosine is positive. These sign requirements immediately tell us which quadrant(s) $\theta$ can lie in.

Let me build a quick picture of the signs:

Quadrant	$\sin\theta$	$\cos\theta$	$\tan\theta = \frac{\sin\theta}{\cos\theta}$
I	$+$	$+$	$+$
II	$+$	$-$	$-$
III	$-$	$-$	$+$
IV	$-$	$+$	$-$

We need $\cos\theta > 0$ (so Quadrant I or IV) and $\tan\theta < 0$ (so Quadrant II or IV). The only quadrant satisfying both is Quadrant IV.

Now let's find the specific angle.

From $\cos\theta = \frac{1}{\sqrt{2}}$ :

The reference angle is $\frac{\pi}{4}$ since $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ . The general solution for $\cos\theta = \frac{1}{\sqrt{2}}$ is:

\theta = 2n\pi \pm \frac{\pi}{4}, \quad n \in \mathbb{Z}

This gives us angles in Quadrant I (when we use $+\frac{\pi}{4}$ ) and Quadrant IV (when we use $-\frac{\pi}{4}$ ).

From $\tan\theta = -1$ :

The reference angle where $|\tan\theta| = 1$ is $\frac{\pi}{4}$ . Since tangent is negative in Quadrants II and IV:

\theta = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z}

For even $n = 2k$ : $\theta = 2k\pi - \frac{\pi}{4}$ (Quadrant IV)

For odd $n = 2k+1$ : $\theta = (2k+1)\pi - \frac{\pi}{4} = 2k\pi + \frac{3\pi}{4}$ (Quadrant II)

Finding the intersection:

From step 1, the Quadrant IV solution is $\theta = 2n\pi - \frac{\pi}{4}$ .

From step 2, the Quadrant IV solution is also $\theta = 2k\pi - \frac{\pi}{4}$ .

These are identical! Both describe the same set of angles. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set A-11 markMCQ
$\cos 5\pi = \cos 4\pi$
›Reveal solutionSolution

Evaluate all four statements using periodicity; only (A) is false, because $\cos 5\pi=-1$ while $\cos 4\pi=+1$ .

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
$\cos 5\pi = \cos(4\pi + \pi) = \cos\pi = -1.$
$4\pi$ is an even multiple of $\pi$ (a full two revolutions), so
$\cos 4\pi = \cos 0 = +1.$
Since $-1 \neq +1$ , this statement is FALSE. This is our candidate.

Step 2 — Option (B): $\sin 2\pi \overset{?}{=} \sin(-2\pi)$ .
$\sin 2\pi = 0, \qquad \sin(-2\pi) = -\sin 2\pi = -0 = 0.$
$0 = 0$ ✓ — 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 $\pi$ :
$\sin 4\pi = 0, \qquad \sin 6\pi = 0.$
✓ — TRUE.

Step 4 — Option (D): $\tan 45^\circ \overset{?}{=} \tan(-315^\circ)$ . …
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KCET 2025Set A-11 markMCQ
$\sec^2(\tan^{-1}2) + \csc^2(\cot^{-1}3) =$
›Reveal solutionSolution

Apply the Pythagorean identities $\sec^2 = 1 + \tan^2$ and $\csc^2 = 1 + \cot^2$ directly to the inverse-function arguments.

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
$\sec^{2}\theta = 1 + \tan^{2}\theta,$ $\sec^{2}(\tan^{-1}2) = 1 + 2^{2} = 1 + 4 = 5.$
Step 2 — Second term. Let $\phi = \cot^{-1}3$ , so $\cot\phi = 3$ (with $\phi \in (0, \pi/2)$ ). Using
$\csc^{2}\phi = 1 + \cot^{2}\phi,$ $\csc^{2}(\cot^{-1}3) = 1 + 3^{2} = 1 + 9 = 10.$
Step 3 — Add.

$5 + 10 = 15.$ …
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COMEDK 2024Set 2024-M1 markMCQ
$\text { If } \sin A=\frac{4}{5} \text { and } \cos B=\frac{-12}{13} \text { where } A \text { and } B \text { lie in first and third quadrant respectively. Then } \cos (A+B)=$
›Reveal solutionSolution
Use $\cos(A+B)=\cos A\cos B-\sin A\sin B$ , first fixing the signs of $\cos A$ and $\sin B$ from the quadrant information. The result is $-\frac{16}{65}$ , which corresponds to option (D).

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

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

$\cos^2 A = 1 - \left(\tfrac{4}{5}\right)^2 = \tfrac{9}{25} \Rightarrow \cos A = \tfrac{3}{5}.$

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

$\sin^2 B = 1 - \left(-\tfrac{12}{13}\right)^2 = \tfrac{25}{169} \Rightarrow \sin B = -\tfrac{5}{13}.$

Apply the formula. …
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COMEDK 2023Set 2023-E1 markMCQ
$\text { If } \operatorname{cosec}(90+A)+x \cos A \cot (90+A)=\sin (90+A) \text { then the value of } x \text { is }$
›Reveal solutionSolution
(Valid wherever sin A is non-zero, which is required for the equation to determine x.)

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: …
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KCET 2022Set C-41 markMCQ
$y= \tan x$
›Reveal solutionSolution
$\dfrac{d}{dx}\tan x = \sec^2 x > 0$ , so $\tan x$ is increasing on every branch; on the II-quadrant branch it runs from $-\infty$ (just right of $\pi/2$ ) up to $0$ (at $\pi$ ).

Step 1 — Fix the interval.

The second quadrant means
$\frac{\pi}{2} < x < \pi .$
Step 2 — Establish the direction (increasing or decreasing) rigorously.

Differentiate:
$\frac{dy}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}.$
A square is never negative, and $\cos x \ne 0$ on the open interval, so
$\sec^2 x > 0 \quad \text{for all } x \in \left(\tfrac{\pi}{2}, \pi\right).$
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

$\lim_{x\to (\pi/2)^{+}} \tan x = -\infty .$

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

$\lim_{x\to \pi^{-}} \tan x = \frac{0}{-1} = 0^{-} .$
Step 4 — Combine. …
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KCET 2021Set A-11 markMCQ
$f(x) = \begin{bmatrix} \cos x & 1 & 0 \\ 0 & 2\cos x & 3 \\ 0 & 1 & 2\cos x \end{bmatrix}$
›Reveal solutionSolution
The determinant of $f(x)$ works out to $4\cos^3 x - 3\cos x$ . Substituting $x=\pi$ (where $\cos\pi=-1$ ) gives the value $\boxed{-1}$ , option (A).

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.

Write the matrix:

$f(x) = \begin{bmatrix} \cos x & 1 & 0 \\ 0 & 2\cos x & 3 \\ 0 & 1 & 2\cos x \end{bmatrix}$

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

$|f(x)| = \cos x \cdot \begin{vmatrix} 2\cos x & 3 \\ 1 & 2\cos x \end{vmatrix}$

Compute the $2\times2$ determinant:

$\begin{vmatrix} 2\cos x & 3 \\ 1 & 2\cos x \end{vmatrix} = (2\cos x)(2\cos x) - (3)(1) = 4\cos^2 x - 3$

Therefore: …
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KCET 2020Set A-11 markMCQ
$\tan A + \cot A = 2$
›Reveal solutionSolution

$\tan A$ and $\cot A$ are reciprocals, and a positive number plus its reciprocal equals $2$ only when the number is $1$ — so both are $1$ .

Step 1 — Reduce to one variable.

Since $\cot A=\dfrac{1}{\tan A}$ , set $t=\tan A$ ( $t\neq0$ ). The condition becomes
$t+\frac{1}{t}=2 .$
Step 2 — Solve.

Multiply through by $t$ :
$t^{2}+1=2t\ \Longrightarrow\ t^{2}-2t+1=0\ \Longrightarrow\ (t-1)^{2}=0\ \Longrightarrow\ t=1 .$
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. …
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←Previous questionQ15 — If sinθ + cosθ = 1, then find the general value of θ.→Next questionQ17 — If cotθ + tanθ = 2cscθ, then find the general value of θ.

Q.Find the most general value of θ\thetaθ satisfying the equation tan⁡θ=−1\tan\theta = -1tanθ=−1 and cos⁡θ=12\cos\theta = \dfrac{1}{\sqrt{2}}cosθ=2​1​.

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Q.Find the most general value of $\theta$ satisfying the equation $\tan\theta = -1$ and $\cos\theta = \dfrac{1}{\sqrt{2}}$ .

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