Q.Find the values of other five trigonometric functions if $\tan x = -\frac{5}{12}$, $x$ lies in second quadrant.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $\tan\alpha = \dfrac{-7}{24}$, $\sec\beta = \dfrac{61}{60}$ and both $\alpha, \beta$ lie in the same quadrant, then $\cos(\alpha+\beta) =$ (A) $\dfrac{1363}{1525}$ (B) $\dfrac{1236}{1525}$ (C) $\dfrac{23}{24}$ (D) $\dfrac{55}{61}$

   ›Reveal solutionSolution

   Identify the common quadrant (QIV, since tangent is negative and secant/cosine is positive there), find $\sin,\cos$ of each angle from Pythagorean triples, and apply the cosine addition formula to get $\dfrac{1363}{1525}$.

   Concept and Intuition

   Given $\tan\alpha$ and $\sec\beta$ (magnitudes only fix the reference triangle; the quadrant fixes the actual signs of $\sin$ and $\cos$). Once both angles' full sine and cosine are pinned down, the compound-angle formula $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ finishes the problem.

   Step-by-Step Solution

   1. $\tan\alpha = -\dfrac{7}{24}$ is negative, and $\sec\beta = \dfrac{61}{60}$ is positive so $\cos\beta>0$. The only quadrant where tangent is negative and cosine is positive is the fourth quadrant (cos $+$, sin $-$, tan $-$). So both $\alpha$ and $\beta$ are in QIV.
   2. For $\alpha$: reference triangle with legs 7, 24, hypotenuse 25 (since $7^2+24^2=49+576=625=25^2$). In QIV, $\cos\alpha = \dfrac{24}{25}$ (positive), $\sin\alpha = -\dfrac{7}{25}$ (negative), consistent with $\tan\alpha=-7/24$.
   3. For $\beta$: $\sec\beta = 61/60 \Rightarrow \cos\beta = 60/61$. Find $\sin\beta$: $\sin^2\beta = 1-\left(\dfrac{60}{61}\right)^2 = \dfrac{61^2-60^2}{61^2}=\dfrac{121}{3721}$, so $|\sin\beta|=\dfrac{11}{61}$. In QIV, sine is negative: $\sin\beta = -\dfrac{11}{61}$. …

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2. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\sin A = \dfrac{3}{5}$ and $A$ lies in the second quadrant, then $\dfrac{\tan A - \sec A}{\cot A + \text{cosec}\, A} =$ (A) $\dfrac{2}{5}$ (B) $\dfrac{11}{15}$ (C) $\dfrac{4}{3}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   This is a straightforward trig-ratio evaluation using quadrant sign rules. Answer: $\dfrac{3}{2}$.

   Concept and Intuition

   In the second quadrant, sine is positive but cosine (and everything derived from it, like tangent, secant) is negative. Once we pin down $\sin A$ and $\cos A$ with correct signs, every other ratio follows directly from their definitions.

   Step-by-Step Solution

   1. Given $\sin A = \dfrac{3}{5}$, and using $\sin^2A+\cos^2A=1$: $\cos^2A = 1-\dfrac{9}{25}=\dfrac{16}{25}$, so $\cos A = \pm\dfrac{4}{5}$.
   2. Since $A$ is in the second quadrant, $\cos A < 0$, so $\cos A = -\dfrac{4}{5}$.
   3. Compute the other ratios:

   $\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{-4/5} = -\frac{3}{4}, \quad \sec A = \frac{1}{\cos A} = -\frac{5}{4}$

   $\cot A = \frac{1}{\tan A} = -\frac{4}{3}, \quad \csc A = \frac{1}{\sin A} = \frac{5}{3}$ …

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3. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $630° < \theta < 810°$ and $\tan\theta = -\dfrac{7}{24}$, then $\cos\left(\dfrac{\theta}{4}\right) =$ (A) $-\sqrt{\dfrac{7+5\sqrt{2}}{10\sqrt{2}}}$ (B) $\sqrt{\dfrac{7+5\sqrt{2}}{2\sqrt{2}}}$ (C) $-\sqrt{\dfrac{5\sqrt{2}-7}{10\sqrt{2}}}$ (D) $\sqrt{\dfrac{5\sqrt{2}-7}{2\sqrt{2}}}$

   ›Reveal solutionSolution

   This is a two-step half-angle problem where correctly tracking the quadrant at each stage is the crux. The answer is (A).

   Concept and Intuition

   Half-angle formulas give $\cos(\phi/2)=\pm\sqrt{\dfrac{1+\cos\phi}{2}}$, and the sign must be fixed by knowing which quadrant $\phi/2$ actually lies in — it cannot be read off from $\phi$'s quadrant alone. Here we must apply the half-angle formula twice (once to get $\theta/2$ from $\theta$, once to get $\theta/4$ from $\theta/2$), pinning the sign each time.

   Step-by-Step Solution

   1. $\tan\theta=-7/24<0$ means $\theta$ (reduced mod $360°$) lies in QII or QIV. The given range $630°<\theta<810°$ reduces (subtract $360°$) to $270°<\theta-360°<450°$, i.e. $\theta-360°$ spans QIV ($270°$–$360°$) then QI ($360°$–$450°$). Only the QIV portion has negative tangent, so $\theta-360°\in(270°,360°)$, i.e. $\theta\in(630°,720°)$.
   2. In that range $\theta$ behaves like a QIV angle: $\cos\theta=24/25>0$, $\sin\theta=-7/25<0$ (the $7$-$24$-$25$ triple).
   3. Then $\theta/2\in(315°,360°)$, which is QIV: $\cos(\theta/2)>0$, $\sin(\theta/2)<0$. $\cos(\theta/2)=\sqrt{\dfrac{1+24/25}{2}}=\sqrt{\dfrac{49/25}{2}}=\dfrac{7}{5\sqrt2}$. …

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4. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\cos\theta=\dfrac{-3}{5}$ and $\theta$ does not lie in second quadrant, then $\tan\dfrac{\theta}{2}=$ (A) 2 (B) 1 (C) -2 (D) -1

   ›Reveal solutionSolution

   This tests quadrant analysis combined with the half-angle tangent formula. Answer: $\tan\frac{\theta}{2}=-2$.

   Concept and Intuition

   Given only $\cos\theta$, there are two possible quadrants (here Q2 or Q3, since cosine is negative in both); the extra condition ("not in Q2") pins down the sign of $\sin\theta$, which then plugs directly into the half-angle formula $\tan\frac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}$.

   Step-by-Step Solution

   1. $\cos\theta=-\dfrac35$ is negative, so $\theta$ lies in Q2 or Q3. Given it's not Q2, $\theta$ is in Q3, where both sine and cosine are negative.
   2. $\sin^2\theta = 1-\cos^2\theta = 1-\dfrac{9}{25}=\dfrac{16}{25}$, so $\sin\theta=\pm\dfrac45$; in Q3, $\sin\theta=-\dfrac45$.
   3. Apply the half-angle identity $\tan\dfrac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}$: $\tan\dfrac{\theta}{2}=\dfrac{-4/5}{1-3/5}=\dfrac{-4/5}{2/5}=-2$. …

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5. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.The general solution satisfying both the equations $\sin x=-\dfrac{3}{5}$ and $\cos x=-\dfrac{4}{5}$ is (A) $x=(2n+1)\pi+\operatorname{Tan}^{-1}\left(\dfrac{3}{4}\right), n\in Z$ (B) $x=2n\pi+\operatorname{Tan}^{-1}\left(\dfrac{3}{4}\right), n\in Z$ (C) $x=n\pi+\operatorname{Tan}^{-1}\left(\dfrac{3}{4}\right), n\in Z$ (D) $x=n\pi\pm\operatorname{Tan}^{-1}\left(\dfrac{3}{4}\right), n\in Z$

   ›Reveal solutionSolution

   This tests finding the general solution when both sine and cosine (not just one) are specified, which pins the angle to one specific quadrant per period. Answer: $x=(2n+1)\pi+\operatorname{Tan}^{-1}\left(\frac34\right)$.

   Concept and Intuition

   When only one of $\sin x$ or $\cos x$ is given, there are two possible quadrants and the general solution has extra branches. But when both are specified simultaneously, exactly one quadrant is consistent every period, so the general solution is a single family $2n\pi+(\text{one specific angle})$ — here re-expressed using odd multiples of $\pi$ since that angle sits in Q3.

   Step-by-Step Solution

   1. Both $\sin x=-\dfrac35$ and $\cos x=-\dfrac45$ are negative $\Rightarrow x$ lies in the third quadrant (Q3).
   2. Find the reference (acute) angle $\alpha$ where $\tan\alpha=\dfrac{|\sin x|}{|\cos x|}=\dfrac{3/5}{4/5}=\dfrac34$, so $\alpha=\operatorname{Tan}^{-1}\left(\dfrac34\right)$.
   3. The Q3 angle with this reference angle is $\pi+\alpha$ (adding $\pi$ to the Q1 reference angle lands in Q3, where both sin and cos flip negative — matches).
   4. The general solution (period $2\pi$) is $x = \pi+\alpha+2n\pi = (2n+1)\pi+\alpha$, $n\in\mathbb Z$ (since $(2n+1)\pi$ ranges over all odd multiples of $\pi$, which is the same set as $\pi+2n\pi$). …

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6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $\cos x+\sin x=\dfrac{1}{2}$ and $0<x<\pi$, then $\tan x=$ (A) $\dfrac{1+\sqrt{7}}{4}$ (B) $\dfrac{1-\sqrt{7}}{4}$ (C) $\dfrac{4-\sqrt{7}}{3}$ (D) $-\dfrac{(4+\sqrt{7})}{3}$

   ›Reveal solutionSolution

   Turn the sum-of-trig equation into a quadratic for $\sin x,\cos x$, pick the branch consistent with $0<x<\pi$, and simplify $\tan x$; the answer is $-\frac{4+\sqrt7}{3}$.

   Concept and Intuition

   Given $\sin x+\cos x=s$, squaring produces $\sin x\cos x$ in terms of $s$, so $\sin x$ and $\cos x$ become the two roots of a quadratic. The range restriction on $x$ then picks out which root is $\sin x$ and which is $\cos x$.

   Step-by-Step Solution

   1. Square: $(\cos x+\sin x)^2=\tfrac14\Rightarrow 1+2\sin x\cos x=\tfrac14\Rightarrow \sin x\cos x=-\tfrac38$.
   2. $\sin x$ and $\cos x$ satisfy $t^2-\left(\tfrac12\right)t-\tfrac38=0$, i.e. $8t^2-4t-3=0$.
   3. Solve: $t=\dfrac{4\pm\sqrt{16+96}}{16}=\dfrac{4\pm\sqrt{112}}{16}=\dfrac{4\pm4\sqrt7}{16}=\dfrac{1\pm\sqrt7}{4}$. …

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7. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $\sin\theta+2\cos\theta=1$ and $\theta$ belongs to $4^{th}$ quadrant (not lying on the coordinate axes) then $7\cos\theta+6\sin\theta=$ (A) $\dfrac{4}{17}$ (B) $2$ (C) $\dfrac{7}{17}$ (D) $\dfrac{4}{5}$

   ›Reveal solutionSolution

   Solve the linear trig equation together with the Pythagorean identity to pin down $\sin\theta,\cos\theta$ in the 4th quadrant; the answer is $2$.

   Concept and Intuition

   A single linear equation in $\sin\theta$ and $\cos\theta$ combined with $\sin^2\theta+\cos^2\theta=1$ gives a quadratic in one variable. The quadrant restriction then selects the valid root.

   Step-by-Step Solution

   1. From $\sin\theta+2\cos\theta=1$: $\sin\theta=1-2\cos\theta$.
   2. Substitute into $\sin^2\theta+\cos^2\theta=1$: $(1-2\cos\theta)^2+\cos^2\theta=1\Rightarrow1-4\cos\theta+4\cos^2\theta+\cos^2\theta=1\Rightarrow5\cos^2\theta-4\cos\theta=0$.
   3. Factor: $\cos\theta(5\cos\theta-4)=0$. Since $\theta$ doesn't lie on an axis, $\cos\theta\ne0$, so $\cos\theta=\tfrac45$.
   4. Then $\sin\theta=1-2\left(\tfrac45\right)=1-\tfrac85=-\tfrac35$. This is consistent with the 4th quadrant ($\sin\theta<0,\cos\theta>0$). …

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8. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $\sec\theta + \tan\theta = \dfrac{1}{3}$, then the quadrant in which $2\theta$ lies is (A) $1^{st}$ quadrant (B) $2^{nd}$ quadrant (C) $3^{rd}$ quadrant (D) $4^{th}$ quadrant

   ›Reveal solutionSolution

   Pair $\sec\theta+\tan\theta$ with $\sec\theta-\tan\theta$ (their product is always 1) to solve for $\theta$'s trig values, then use double-angle formulas to locate $2\theta$. Answer: 3rd quadrant.

   Concept and Intuition

   The identity $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^2\theta-\tan^2\theta=1$ lets you find $\sec\theta-\tan\theta$ immediately once $\sec\theta+\tan\theta$ is known, without ambiguity — a much faster route than solving a trig equation directly.

   Step-by-Step Solution

   1. Given $\sec\theta+\tan\theta=\dfrac13$. Since $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$, we get $\sec\theta-\tan\theta=3$.
   2. Adding: $2\sec\theta = \dfrac13+3=\dfrac{10}3 \Rightarrow \sec\theta=\dfrac53 \Rightarrow \cos\theta=\dfrac35$.
   3. Subtracting: $2\tan\theta=\dfrac13-3=-\dfrac83 \Rightarrow \tan\theta=-\dfrac43$.
   4. $\sin\theta=\tan\theta\cdot\cos\theta = -\dfrac43\cdot\dfrac35=-\dfrac45$ (check: $(3/5)^2+(4/5)^2=1$ ✓). So $\cos\theta>0,\sin\theta<0$: $\theta$ is in Q4. …

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9. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $540^\circ < A < 630^\circ$ and $|\cos A| = \dfrac{5}{13}$ then $\tan\dfrac{A}{2}\tan A =$ (A) $\dfrac{18}{5}$ (B) $\dfrac{8}{5}$ (C) $-\dfrac{8}{5}$ (D) $-\dfrac{18}{5}$

   ›Reveal solutionSolution

   Locate the quadrant of $A$ from the given range, fix the sign of $\cos A$ and hence $\sin A$, then compute $\tan(A/2)$ via the half-angle formula. Answer: $-18/5$.

   Concept and Intuition

   The range $540^\circ$ to $630^\circ$ is just the 3rd quadrant shifted by a full $360^\circ$ rotation, so all trig signs behave exactly as in $180^\circ$–$270^\circ$. The half-angle formula $\tan(A/2)=\dfrac{\sin A}{1+\cos A}$ avoids ambiguity about which root/sign to take for $A/2$.

   Step-by-Step Solution

   1. $540^\circ<A<630^\circ \Rightarrow 180^\circ<A-360^\circ<270^\circ$: 3rd quadrant, so $\cos A<0,\ \sin A<0$.
   2. $|\cos A|=5/13 \Rightarrow \cos A=-5/13$. Since $\sin^2A=1-25/169=144/169$ and $\sin A<0$: $\sin A=-12/13$.
   3. $\tan\dfrac A2 = \dfrac{\sin A}{1+\cos A}=\dfrac{-12/13}{1-5/13}=\dfrac{-12/13}{8/13}=-\dfrac{12}{8}=-\dfrac32$.
   4. $\tan A = \dfrac{\sin A}{\cos A}=\dfrac{-12/13}{-5/13}=\dfrac{12}{5}$. …

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10. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.If $\alpha$ is in the 3rd Quadrant, $\beta$ is in the 2nd Quadrant such that $\tan\alpha = \dfrac{1}{7}$, $\sin\beta = \dfrac{1}{\sqrt{10}}$, then $\sin(2\alpha+\beta) =$ (A) $\dfrac{3 \times \sqrt{10}}{25}$ (B) $\dfrac{3}{\sqrt{10}}$ (C) $\dfrac{3}{25\sqrt{10}}$ (D) $\dfrac{\sqrt{10}}{3 \times 25}$

    ›Reveal solutionSolution

    The key idea is to compute $\sin(2\alpha+\beta)$ using the sine addition formula, which requires $\sin\alpha$, $\cos\alpha$, $\sin\beta$, and $\cos\beta$. Using quadrant signs and given values, we find $\sin(2\alpha+\beta) = \frac{3}{25\sqrt{10}}$, matching option (C).

    We are given:

    • $\alpha$ in Quadrant III, $\tan\alpha = \frac{1}{7}$
    • $\beta$ in Quadrant II, $\sin\beta = \frac{1}{\sqrt{10}}$ We need $\sin(2\alpha+\beta)$.

    ---

    Concept and Intuition

    The formula $\sin(2\alpha+\beta) = \sin 2\alpha \cos\beta + \cos 2\alpha \sin\beta$ is our path.

    To use it, we need $\sin\alpha$, $\cos\alpha$, $\sin\beta$, $\cos\beta$ — but with correct signs based on quadrants.

    Quadrant III: both sine and cosine are negative.

    Quadrant II: sine positive, cosine negative.

    Getting these signs wrong is the classic pitfall.

    ---

    Step-by-step solution

    1. Find $\sin\alpha$ and $\cos\alpha$ from $\tan\alpha = \frac{1}{7}$ in Quadrant III $\tan\alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{7}$ means a right triangle with legs 1 and 7, hypotenuse $\sqrt{1^2+7^2} = \sqrt{50} = 5\sqrt{2}$. In Quadrant III, both sine and cosine are negative:

    $\sin\alpha = -\frac{1}{5\sqrt{2}}, \quad \cos\alpha = -\frac{7}{5\sqrt{2}}.$

    2. Find $\cos\beta$ from $\sin\beta = \frac{1}{\sqrt{10}}$ in Quadrant II $\sin\beta = \frac{1}{\sqrt{10}}$ gives a triangle with opposite 1, hypotenuse $\sqrt{10}$, so adjacent = $\sqrt{10-1} = 3$. In Quadrant II, cosine is negative:

    $\cos\beta = -\frac{3}{\sqrt{10}}.$

    3. Compute $\sin 2\alpha$ and $\cos 2\alpha$ Use double-angle formulas:

    $\sin 2\alpha = 2\sin\alpha\cos\alpha = 2\left(-\frac{1}{5\sqrt{2}}\right)\left(-\frac{7}{5\sqrt{2}}\right) = 2 \cdot \frac{7}{25 \cdot 2} = \frac{7}{25}.$

    $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \left(\frac{49}{50}\right) - \left(\frac{1}{50}\right) = \frac{48}{50} = \frac{24}{25}.$

    (Note: $\cos 2\alpha$ is positive; $\alpha$ in QIII means $2\alpha$ could be in QI or QII, but here it’s QI.)

    4. Apply the sine addition formula $$$$

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If $\sin\theta=\dfrac{3}{5}$ and $\theta$ is not in the first quadrant, then $15\sin2\theta-20\cos2\theta-7\tan2\theta=$ (A) $-4$ (B) $-12$ (C) $12$ (D) $4$

    ›Reveal solutionSolution

    Fixing the correct quadrant for $\theta$ (Q2, since $\sin\theta>0$ but $\theta\notin$ Q1) determines the sign of $\cos\theta$, and from there all the double-angle values follow. Answer: 4.

    Concept and Intuition

    $\sin\theta=3/5$ alone doesn't fix $\theta$'s quadrant — it could be Q1 or Q2 since sine is positive in both. The extra condition "$\theta$ is not in the first quadrant" forces Q2, where cosine is negative. Getting this sign right is the crux of the whole problem.

    Step-by-Step Solution

    1. Since $\sin\theta=3/5>0$ and $\theta$ is not in Q1, $\theta$ must be in Q2, so $\cos\theta=-\sqrt{1-9/25}=-4/5$.
    2. $\sin2\theta=2\sin\theta\cos\theta=2\cdot\frac35\cdot\left(-\frac45\right)=-\frac{24}{25}$.
    3. $\cos2\theta=1-2\sin^2\theta=1-\frac{18}{25}=\frac{7}{25}$.
    4. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{-24/25}{7/25}=-\dfrac{24}{7}$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Which of the following trigonometric values are negative? I) $\sin(-292^\circ)$ II) $\tan(-193^\circ)$ III) $\cos(-207^\circ)$ IV) $\cot(-222^\circ)$ (A) II, III and IV (B) III only (C) I and III (D) II and III

    ›Reveal solutionSolution

    Add $360^\circ$ to each angle: $68^\circ$ (Q1), $167^\circ$, $153^\circ$, $138^\circ$ (all Q2). In Q2 only sine/cosec are positive — so $\tan,\cos,\cot$ are negative: II, III, IV.

    Concept and Intuition

    Two facts settle every question of this type.

    1. Coterminal angles: $\theta$ and $\theta+360^\circ$ have the same terminal ray, hence identical values of all six functions. So a negative angle is best replaced by $\theta+360^\circ$.
    2. ASTC (All–Sin–Tan–Cos): in Q1 all are $+$; in Q2 only $\sin$ (and $\csc$); in Q3 only $\tan$ (and $\cot$); in Q4 only $\cos$ (and $\sec$).

    Step-by-Step Solution

    1. I) $\sin(-292^\circ)$: $-292^\circ+360^\circ=68^\circ$, first quadrant. All functions positive there, so $\sin(-292^\circ)=\sin 68^\circ>0$. Not negative.
    2. II) $\tan(-193^\circ)$: $-193^\circ+360^\circ=167^\circ$, second quadrant. In Q2 tangent is negative: $\tan 167^\circ<0$. Negative.
    3. III) $\cos(-207^\circ)$: $-207^\circ+360^\circ=153^\circ$, second quadrant. In Q2 cosine is negative. Negative.
    4. IV) $\cot(-222^\circ)$: $-222^\circ+360^\circ=138^\circ$, second quadrant. Cotangent shares the sign of tangent, which is negative in Q2. Negative.
    5. Hence the negative ones are II, III and IV. …

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