Q.Write $\cot^{-1}\left(\dfrac{1}{\sqrt{x^{2}-1}}\right)$, $x > 1$ in the simplest form.

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A} =$ (A) $\sec A \csc A - 1$ (B) $\tan A + \cot A$ (C) $\tan A + \cot A + 1$ (D) $\sec A + \csc A + 1$

   ›Reveal solutionSolution

   Substituting $t=\tan A$ turns the expression into an algebraic fraction that factors via $t^3-1$, simplifying cleanly to $\tan A+\cot A+1$.

   Concept and Intuition

   Trig identities involving $\tan$ and $\cot$ of the same angle often simplify beautifully after substituting $t=\tan A$ (so $\cot A = 1/t$), turning the problem into ordinary algebra, and using the factorization $t^3-1=(t-1)(t^2+t+1)$.

   Step-by-Step Solution

   1. Let $t=\tan A$, so $\cot A=\dfrac1t$.
   2. First term: $\dfrac{\tan A}{1-\cot A}=\dfrac{t}{1-\frac1t}=\dfrac{t}{\frac{t-1}{t}}=\dfrac{t^2}{t-1}$.
   3. Second term: $\dfrac{\cot A}{1-\tan A}=\dfrac{1/t}{1-t}=\dfrac{1}{t(1-t)}=\dfrac{-1}{t(t-1)}$.
   4. Sum $=\dfrac{t^2}{t-1}-\dfrac{1}{t(t-1)}=\dfrac{t^3-1}{t(t-1)}$. …

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2. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $1-\cot 23^0 = \dfrac{x}{1-\cot 22^0}$, then x = (A) 1 (B) 2 (C) $\dfrac{1}{2}$ (D) 3

   ›Reveal solutionSolution

   Recognizing $23^\circ+22^\circ=45^\circ$ triggers a standard cotangent identity that instantly evaluates the product without needing actual trig values.

   Concept and Intuition

   Whenever $A+B=45^\circ$, the identity $(\cot A-1)(\cot B-1)=2$ holds — this follows directly from the cotangent addition formula $\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1$.

   Step-by-Step Solution

   1. Rearrange the given equation: $x=(1-\cot23^\circ)(1-\cot22^\circ)$.
   2. Note $23^\circ+22^\circ=45^\circ$.
   3. From $\cot(A+B)=1$ when $A+B=45^\circ$: $\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1 \Rightarrow \cot A\cot B-1=\cot A+\cot B$.
   4. So $\cot A\cot B-\cot A-\cot B+1 = (\cot A+\cot B+1)-\cot A-\cot B+1=2$, i.e. $(\cot A-1)(\cot B-1)=2$. …

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3. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.What is the value of $\cos\left(22\dfrac{1}{2}\right)^{\circ} =$ (A) $\sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}$ (B) $\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}$ (C) $\sqrt{2} - 1$ (D) $\sqrt{2} + 1$

   ›Reveal solutionSolution

   Applying the half-angle cosine formula to $45^\circ$ and simplifying matches option (B) after rationalizing.

   Concept and Intuition

   $22.5^\circ$ is half of $45^\circ$, a standard angle. The half-angle identity $\cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{1+\cos x}{2}}$ (positive root since $22.5^\circ$ is in the first quadrant) converts an unfamiliar angle into known values of $\cos45^\circ$.

   Step-by-Step Solution

   1. Write $22.5^\circ = \dfrac{45^\circ}{2}$.
   2. Apply the half-angle formula: $\cos(22.5^\circ) = \sqrt{\dfrac{1+\cos45^\circ}{2}}$.
   3. Substitute $\cos45^\circ = \dfrac{\sqrt2}{2}$: $\cos(22.5^\circ) = \sqrt{\dfrac{1+\frac{\sqrt2}{2}}{2}} = \sqrt{\dfrac{2+\sqrt2}{4}}$.
   4. Rationalize option (B): $\dfrac{\sqrt2+1}{2\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2} = \dfrac{2+\sqrt2}{4}$ — identical to step 3. …

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4. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.$\cot 18^\circ \cdot \cot 36^\circ + 1 =$ (A) $\sqrt{5+2\sqrt5}$ (B) $\sqrt{5-2\sqrt5}$ (C) $3-\sqrt5$ (D) $3+\sqrt5$

   ›Reveal solutionSolution

   $\cot18^\circ\cot36^\circ+1$ numerically equals $5.236\ldots$, which is exactly $3+\sqrt5$.

   Concept and Intuition

   $18^\circ$ and $36^\circ$ are the classic "golden ratio" angles from the regular pentagon, whose trig values involve $\sqrt5$ (e.g. $\cos36^\circ=\frac{1+\sqrt5}{4}$, $\sin18^\circ=\frac{\sqrt5-1}{4}$ — up to the standard forms). Rather than re-deriving these surds from scratch, a clean way to confirm the answer is to numerically evaluate both sides and match against the surd-valued options, since only one can agree to several decimal places.

   Step-by-Step Solution

   1. $\tan18^\circ\approx0.32492\Rightarrow\cot18^\circ\approx3.07768$.
   2. $\tan36^\circ\approx0.72654\Rightarrow\cot36^\circ\approx1.37638$.
   3. Product: $3.07768\times1.37638\approx4.2360$. Adding $1$: $\approx5.2360$. …

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5. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\cos\theta(\csc\theta - \sec\theta) - \cot\theta =$ (A) -1 (B) 1 (C) 0 (D) $\cos^2\theta - \tan^2\theta$

   ›Reveal solutionSolution

   Direct expansion into sines and cosines shows the $\cot\theta$ terms cancel, leaving $-1$.

   Concept and Intuition

   When an expression mixes several different trig ratios ($\csc,\sec,\cot$), it's usually fastest to

   rewrite everything in terms of $\sin\theta$ and $\cos\theta$ and simplify directly, rather than

   hunting for a named identity.

   Step-by-Step Solution

   1. Expand: $\cos\theta(\csc\theta-\sec\theta) = \cos\theta\csc\theta - \cos\theta\sec\theta$.
   2. $\cos\theta\csc\theta = \cos\theta\cdot\dfrac{1}{\sin\theta} = \dfrac{\cos\theta}{\sin\theta}=\cot\theta$.
   3. $\cos\theta\sec\theta = \cos\theta\cdot\dfrac1{\cos\theta}=1$.
   4. So the expression becomes $\cot\theta - 1 - \cot\theta = -1$.

   Common Mistakes …

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6. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$2\cot^2\theta - \cot\theta - 3 =$ (A) $(2\cot\theta - 3)(\cot\theta + 1)$ (B) $(2\cot\theta - 1)(\cot\theta + 3)$ (C) $(2\cot\theta + 3)(\cot\theta - 1)$ (D) $(2\cot\theta + 1)(\cot\theta - 3)$

   ›Reveal solutionSolution

   A plain quadratic factorisation in $\cot\theta$: $2\cot^2\theta-\cot\theta-3=(2\cot\theta-3)(\cot\theta+1)$.

   Concept and Intuition

   Trig "identities" that are really just quadratics in disguise are best handled by substituting a

   single variable for the trig function, factoring as ordinary algebra, then substituting back.

   Step-by-Step Solution

   1. Let $c=\cot\theta$. The expression is $2c^2-c-3$.
   2. Find factors of $2\times(-3)=-6$ that sum to the middle coefficient $-1$: these are $2$ and $-3$.
   3. Split the middle term: $2c^2+2c-3c-3 = 2c(c+1)-3(c+1) = (2c-3)(c+1)$.
   4. Substitute back: $(2\cot\theta-3)(\cot\theta+1)$. …

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7. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.The value of $\tan\left(\dfrac{7\pi}{8}\right)$ is (A) $\sqrt{2}-1$ (B) $1-\sqrt{2}$ (C) $1+\sqrt{2}$ (D) $\dfrac{1}{1+\sqrt{2}}$

   ›Reveal solutionSolution

   Using $\tan(\pi-\theta)=-\tan\theta$ and the known value $\tan\frac{\pi}{8}=\sqrt2-1$, we get $\tan\frac{7\pi}{8}=1-\sqrt2$.

   Concept and Intuition

   Angles in the second quadrant can always be reduced to a first-quadrant reference angle using supplementary-angle identities; here $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$.

   Step-by-Step Solution

   1. Write $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$.
   2. Use $\tan(\pi-\theta) = -\tan\theta$, so $\tan\frac{7\pi}{8} = -\tan\frac{\pi}{8}$.
   3. Recall (or derive from half-angle formula with $\theta=\pi/4$): $\tan\frac{\pi}{8} = \tan22.5^\circ = \sqrt2-1$. …

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8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $\cot x \cot y = a$ and $x + y = \dfrac{\pi}{6}$, then the quadratic equation satisfying $\cot x$ and $\cot y$ is (A) $t^2 + (1-a)\sqrt{3}\,t + a = 0$ (B) $\sqrt{3}t^2 + (1-a)t + a\sqrt{3} = 0$ (C) $\sqrt{3}t^2 + (a-1)t + a\sqrt{3} = 0$ (D) $t^2 + (a-1)\sqrt{3}\,t + a = 0$

   ›Reveal solutionSolution

   Using the cotangent-addition formula with $x+y=\pi/6$ gives the sum $\cot x+\cot y$ in terms of $a$; combined with the product $a$, Vieta's formulas build the quadratic. Answer: (B).

   Concept and Intuition

   Knowing the sum and product of two quantities lets us write the monic quadratic $t^2-(\text{sum})t+(\text{product})=0$ with exactly those roots.

   Step-by-Step Solution

   1. $\cot(x+y) = \dfrac{\cot x\cot y - 1}{\cot y+\cot x}$.
   2. $x+y=\pi/6 \Rightarrow \cot(x+y)=\sqrt3$.
   3. $\cot x\cot y = a$. So $\sqrt3 = \dfrac{a-1}{\cot x+\cot y} \Rightarrow \cot x+\cot y = \dfrac{a-1}{\sqrt3}$.
   4. Sum $S=\dfrac{a-1}{\sqrt3}$, product $P=a$: quadratic is $t^2 - St + P = 0$. …

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9. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.$1 + \cot^2 30^0 - \sec^2 45^0 =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1-\sqrt{3}}{2}$ (C) $2$ (D) $0$

   ›Reveal solutionSolution

   Direct substitution of standard angle values gives $1+3-2=2$.

   Concept and Intuition

   This is a pure standard-angle evaluation: $\cot30°=\sqrt3$ and $\sec45°=\sqrt2$ are memorized exact values.

   Step-by-Step Solution

   1. $\cot30° = \sqrt3 \Rightarrow \cot^230°=3$.
   2. $\sec45° = \sqrt2 \Rightarrow \sec^245°=2$.
   3. Substitute: $1+3-2=2$. …

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10. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If $1+\sqrt{1+a}=(1+\sqrt{1-a})\cot\alpha$ and $0<a<1$, then $\sin 4\alpha=$ (A) $a$ (B) $2a$ (C) $3a$ (D) $4a$

    ›Reveal solutionSolution

    A clever substitution $a=\sin2\theta$ converts the given surd equation into a clean tangent-addition identity, showing directly that $\sin4\alpha=a$.

    Concept and Intuition

    Expressions like $\sqrt{1\pm\sin2\theta}$ simplify beautifully to $|\cos\theta\pm\sin\theta|$ because $1\pm\sin2\theta=(\sin\theta\pm\cos\theta)^2$. This is the standard trick for equations dressed up with nested square roots of $1\pm a$.

    Step-by-Step Solution

    1. Since $0<a<1$, write $a=\sin2\theta$ with $\theta\in(0,\pi/4)$.
    2. $1+a=1+\sin2\theta=(\sin\theta+\cos\theta)^2\Rightarrow\sqrt{1+a}=\sin\theta+\cos\theta$.
    3. $1-a=1-\sin2\theta=(\cos\theta-\sin\theta)^2\Rightarrow\sqrt{1-a}=\cos\theta-\sin\theta$ (positive since $\theta<\pi/4$).
    4. Given equation: $1+\sin\theta+\cos\theta=(1+\cos\theta-\sin\theta)\cot\alpha$.
    5. Using half-angle forms $1+\cos\theta=2\cos^2(\theta/2)$ and $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$: $\cot\alpha=\dfrac{2\cos(\theta/2)[\cos(\theta/2)+\sin(\theta/2)]}{2\cos(\theta/2)[\cos(\theta/2)-\sin(\theta/2)]}=\dfrac{1+\tan(\theta/2)}{1-\tan(\theta/2)}=\tan\!\left(\dfrac\pi4+\dfrac\theta2\right).$ …

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11. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.In a triangle $ABC$, suppose none on the angles are multiples of $\dfrac{\pi}{2}$, then what is the value $\cot A\cot B + \cot B\cot C + \cot A\cot C =$ (A) $\infty$ (B) 1 (C) -1 (D) 0

    ›Reveal solutionSolution

    This is the standard triangle identity $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$, valid whenever $A+B+C=\pi$.

    Concept and Intuition

    Whenever three angles sum to $\pi$, several "conjugate" identities hold between their trig functions — the most famous being $\tan A+\tan B+\tan C=\tan A\tan B\tan C$. The cotangent identity here is the natural analogue, derivable from $C=\pi-(A+B)$.

    Step-by-Step Solution

    1. Since $C=\pi-(A+B)$, $\cot C=-\cot(A+B)=-\dfrac{\cot A\cot B-1}{\cot A+\cot B}=\dfrac{1-\cot A\cot B}{\cot A+\cot B}$.
    2. So $\cot C(\cot A+\cot B)=1-\cot A\cot B$.
    3. Expand: $\cot A\cot C+\cot B\cot C=1-\cot A\cot B$.
    4. Rearranging: $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$.

    Common Mistakes …

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12. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.Given $\frac{\sin 1^\circ}{\sin x^\circ \sin(x+1)^\circ} = \cot x^\circ - \cot(x+1)^\circ$, then the value of $\frac{1}{\sin45^\circ \sin46^\circ} + \frac{1}{\sin46^\circ \sin47^\circ} + \cdots + \frac{1}{\sin89^\circ \sin90^\circ}$ is (A) $\sin 1^\circ$ (B) $\cot 1^\circ$ (C) $-\cot 1^\circ$ (D) $\operatorname{cosec} 1^\circ$

    ›Reveal solutionSolution

    This is a telescoping cotangent sum; it collapses to $\csc1^\circ$.

    Concept and Intuition

    The given identity converts each reciprocal-sine-product term into a difference of cotangents, so consecutive terms cancel out (telescoping), leaving only the first and last cotangent values.

    Step-by-Step Solution

    1. From the given identity, $\frac{1}{\sin x^\circ\sin(x+1)^\circ} = \frac{\cot x^\circ-\cot(x+1)^\circ}{\sin1^\circ}$.
    2. Sum for $x=45,46,\dots,89$:

    $\sum_{x=45}^{89}\frac{1}{\sin x^\circ\sin(x+1)^\circ} = \frac{1}{\sin1^\circ}\sum_{x=45}^{89}\big(\cot x^\circ-\cot(x+1)^\circ\big)$

    3. The sum telescopes to $\cot45^\circ-\cot90^\circ = 1-0=1$. …

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