Q.Show that $\cos\left(2\tan^{-1}\frac{1}{7}\right)=\sin\left(4\tan^{-1}\frac{1}{3}\right)$.

Showing the 12 most recent of 15 on this concept.

1. COMEDK 2026Set 2026-M1 markMCQ
   Q.Which of the following is the simplest form of the expression $\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\sqrt{\mathbf{1 + x ^ { \mathbf { 2 } }}}-\mathbf{1}}{\boldsymbol{x}}\right)$ where $x \neq 0$ (A) $2 \tan ^{-1} x$ (B) $\tan ^{-1} \frac{x}{2}$ (C) $\frac{1}{2} \tan ^{-1} x$ (D) $\tan ^{-1} x$

   ›Reveal solutionSolution

   The expression simplifies to $\frac{1}{2}\tan^{-1}x$ by substituting $x = \tan\theta$ and using trigonometric identities. The correct option is (C).

   We start with the expression

   $\tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right),\quad x\neq 0.$

   The presence of $\sqrt{1+x^2}$ strongly suggests a trigonometric substitution: let $x = \tan\theta$, because then $1+x^2 = 1+\tan^2\theta = \sec^2\theta$, and $\sqrt{1+x^2} = |\sec\theta|$. For simplicity, we can assume $\theta$ in a range where $\sec\theta > 0$ (e.g., $-\pi/2 < \theta < \pi/2$), which covers all real $x$ since $\tan$ is onto $\mathbb{R}$ on that interval.

   1. Substitute $x = \tan\theta$ Then $\sqrt{1+x^2} = \sec\theta$, and the expression becomes

   $\tan^{-1}\left(\frac{\sec\theta - 1}{\tan\theta}\right).$

   2. Rewrite in terms of sine and cosine

   $\frac{\sec\theta - 1}{\tan\theta} = \frac{\frac{1}{\cos\theta} - 1}{\frac{\sin\theta}{\cos\theta}} = \frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1-\cos\theta}{\sin\theta}.$

   3. Use a half-angle identity Recall the identity:

   $\frac{1-\cos\theta}{\sin\theta} = \tan\frac{\theta}{2}.$

   (Derivation: $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$, $1-\cos\theta = 2\sin^2\frac{\theta}{2}$, so the ratio is $\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}$.)

   4. Simplify the inverse tangent So we have $\tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2},$ …

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2. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=$

   (A) tan 54$^\circ$ (B) tan 36$^\circ$ (C) tan 18$^\circ$ (D) tan 9$^\circ$

   ›Reveal solutionSolution

   The expression simplifies to $\tan 54^\circ$ by rewriting the numerator and denominator using sine/cosine of complementary angles and applying the tangent addition formula. The correct option is (A).

   We start with the expression

   $\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ}.$

   The key idea is to transform this into a form that matches the tangent of a sum or difference. Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we want the numerator and denominator to look like $\sin(A+B)$ and $\cos(A+B)$ or to directly yield a tangent ratio.

   1. Rewrite $\sin 9^\circ$ as a cosine of a complementary angle Recall $\sin \theta = \cos(90^\circ - \theta)$. So

   $\sin 9^\circ = \cos(81^\circ).$

   Then the expression becomes

   $\frac{\cos 9^\circ + \cos 81^\circ}{\cos 9^\circ - \cos 81^\circ}.$

   2. Use sum-to-product identities The formulas:

   $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2},$

   $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}.$

   Here $A = 9^\circ$, $B = 81^\circ$. Then

   $\frac{A+B}{2} = \frac{90^\circ}{2} = 45^\circ, \quad \frac{A-B}{2} = \frac{9^\circ - 81^\circ}{2} = \frac{-72^\circ}{2} = -36^\circ.$

   So

   $\cos 9^\circ + \cos 81^\circ = 2\cos 45^\circ \cos(-36^\circ) = 2\cdot\frac{\sqrt{2}}{2}\cdot\cos 36^\circ = \sqrt{2}\cos 36^\circ,$

   and

   $\cos 9^\circ - \cos 81^\circ = -2\sin 45^\circ \sin(-36^\circ) = -2\cdot\frac{\sqrt{2}}{2}\cdot(-\sin 36^\circ) = \sqrt{2}\sin 36^\circ.$

   3. Form the ratio The original expression becomes $$$$

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3. COMEDK 2024Set 2024-M1 markMCQ
   Q.$\text { The value of } \sin ^{-1}\left[\cot \left(\frac{1}{2} \tan ^{-1} \frac{1}{\sqrt{3}}+\cos ^{-1} \frac{\sqrt{12}}{4}+\sin ^{-1} \frac{1}{\sqrt{2}}\right)\right]$ is (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{4}$ (D) 0

   ›Reveal solutionSolution

   The expression simplifies to $\sin^{-1}(\cot(\pi/2)) = \sin^{-1}(0) = 0$, so the correct option is (D).

   We start with a messy nested inverse trigonometric expression. The key is to simplify from the inside out: evaluate each inverse trig term, combine them, then take the cotangent of half an arctangent, and finally apply the outer arcsine. The trickiest part is the half-angle formula for tangent, which lets us handle $\frac{1}{2}\tan^{-1}\frac{1}{\sqrt{3}}$ neatly.

   1. Simplify the known inverse trig values

      • $\tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$, because $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$.
      • $\cos^{-1}\frac{\sqrt{12}}{4}$: note $\sqrt{12} = 2\sqrt{3}$, so $\frac{\sqrt{12}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$. Thus $\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$.
      • $\sin^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4}$, since $\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

   2. Combine the angles inside the cotangent

      The argument of the cotangent is

   $\frac{1}{2}\cdot\frac{\pi}{6} + \frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi}{12} + \frac{\pi}{6} + \frac{\pi}{4}.$

   Get a common denominator of 12:

   $\frac{\pi}{12} + \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}.$

   So the expression inside the outer arcsine becomes $\cot\left(\frac{\pi}{2}\right)$.

   3. Evaluate the cotangent $\cot\frac{\pi}{2} = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0$. …

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4. KCET 2021Set A-11 markMCQ
   Q.$\cos \left[ \cot^{-1}(-\sqrt{3}) + \dfrac{\pi}{6} \right] =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{\sqrt{2}}$ (D) $-1$

   ›Reveal solutionSolution

   Evaluate the inverse cotangent in its principal range $(0,\pi)$ — it is $5\pi/6$, not $-\pi/6$ — then the bracket becomes $\pi$.

   Step 1 — The principal-value branch (this is the whole trap).

   For $\cot^{-1}x$ the principal range is

   $\cot^{-1}:\ \mathbb{R}\to(0,\pi).$

   Unlike $\tan^{-1}$, it is never negative. So a negative argument gives an answer in the second quadrant, using

   $\cot^{-1}(-x)=\pi-\cot^{-1}(x),\qquad x>0 .$

   Step 2 — Evaluate $\cot^{-1}(-\sqrt3)$.

   First, $\cot\frac{\pi}{6}=\dfrac{\cos(\pi/6)}{\sin(\pi/6)}=\dfrac{\sqrt3/2}{1/2}=\sqrt3$, so $\cot^{-1}(\sqrt3)=\dfrac{\pi}{6}$.

   Therefore

   $\cot^{-1}(-\sqrt3)=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$

   Check: $\cot\frac{5\pi}{6}=\dfrac{-\sqrt3/2}{1/2}=-\sqrt3$ ✓, and $\frac{5\pi}{6}\in(0,\pi)$ ✓. …

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5. COMEDK 2025Set 2025-A1 markMCQ
   Q.If $2 y=\left[\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right]^2 \forall x \in\left(0, \frac{\pi}{2}\right)$ then $\frac{d y}{d x}$ is equal to : (A) $x-\frac{\pi}{6}$ (B) $2 x-\frac{\pi}{3}$ (C) $\frac{\pi}{6}-x$ (D) $\frac{\pi}{3}-x$

   ›Reveal solutionSolution

   Simplifying the $\cot^{-1}$ argument gives $u=\cot^{-1}(\text{arg})=x-\frac{\pi}{6}$, so $\frac{dy}{dx}=u=x-\frac{\pi}{6}$.

   Write the argument using compound angles. Dividing through by $2$:

   $\sqrt{3}\cos x+\sin x = 2\cos\!\left(x-\tfrac{\pi}{6}\right),\qquad \cos x-\sqrt{3}\sin x = 2\cos\!\left(x+\tfrac{\pi}{3}\right).$

   Since $x+\frac{\pi}{3}=\left(x-\frac{\pi}{6}\right)+\frac{\pi}{2}$, we have $\cos\!\left(x+\tfrac{\pi}{3}\right)=-\sin\!\left(x-\tfrac{\pi}{6}\right)$. Hence

   $\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}=\frac{\cos\!\left(x-\frac{\pi}{6}\right)}{-\sin\!\left(x-\frac{\pi}{6}\right)}=-\cot\!\left(x-\tfrac{\pi}{6}\right)=\cot\!\left(\tfrac{\pi}{6}-x\right).$ …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)\left(\tan \frac{\pi}{12}+\cot \frac{\pi}{12}\right)=$

   (A) $\sqrt{2}$ (B) $\frac{1}{\sqrt{2}}$ (C) $4 \sqrt{2}$ (D) $2 \sqrt{2}$

   ›Reveal solutionSolution

   The expression simplifies to $2\sqrt{2}$ by rewriting $\tan$ and $\cot$ in terms of sine and cosine, combining them into a single fraction, and then using the double-angle identity for sine. The correct option is (D).

   ---

   The key insight is that $\tan$ and $\cot$ are reciprocals, so their sum can be expressed as a single fraction with a common denominator. That denominator will be $\sin x \cos x$, which is exactly $\frac{1}{2}\sin 2x$. Meanwhile, the first factor $\cos x - \sin x$ can be paired with the numerator of that fraction to produce a neat cancellation or simplification using known exact values.

   Let’s set $x = \frac{\pi}{12}$ to keep notation clean.

   1. Rewrite the second factor

   $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}.$

   So the whole expression becomes

   $(\cos x - \sin x) \cdot \frac{1}{\sin x \cos x}.$

   2. Combine into a single fraction

   $\frac{\cos x - \sin x}{\sin x \cos x}.$

   3. Use the double-angle identity Recall $\sin 2x = 2\sin x \cos x$, so $\sin x \cos x = \frac{1}{2}\sin 2x$. Then

   $\frac{\cos x - \sin x}{\frac{1}{2}\sin 2x} = \frac{2(\cos x - \sin x)}{\sin 2x}.$

   4. Simplify the numerator

      Notice $\cos x - \sin x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x\right) = \sqrt{2}\cos\left(x + \frac{\pi}{4}\right)$.

      But here it’s easier: for $x = \frac{\pi}{12}$, we have $2x = \frac{\pi}{6}$.

      So $\sin 2x = \sin\frac{\pi}{6} = \frac{1}{2}$.

   5. Evaluate $\cos x - \sin x$ exactly

   $\cos\frac{\pi}{12} - \sin\frac{\pi}{12}.$

   Using known values: $\cos\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$, $\sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$.

   Their difference:

   $$$$

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7. KCET 2018Set A-11 markMCQ
   Q.The value of the expression $\tan\left(\frac{1}{2} \cos^{-1} \frac{2}{\sqrt{5}}\right)$ is (A) $2 - \sqrt{5}$ (B) $\sqrt{5} - 2$ (C) $\frac{\sqrt{5} - 2}{2}$ (D) $5 - \sqrt{2}$

   ›Reveal solutionSolution

   Set $\theta = \cos^{-1}\frac{2}{\sqrt5}$, find $\sin\theta$ from the Pythagorean identity, then use the half-angle identity $\tan\frac\theta2 = \frac{1-\cos\theta}{\sin\theta}$.

   Step 1 — Introduce the angle.

   Let

   $\theta = \cos^{-1}\frac{2}{\sqrt5} \quad\Longrightarrow\quad \cos\theta = \frac{2}{\sqrt5},\qquad \theta \in \left[0, \tfrac{\pi}{2}\right]$

   (the principal value is in the first quadrant since $\frac{2}{\sqrt5} > 0$). The expression we want is $\tan\dfrac{\theta}{2}$.

   Step 2 — Get $\sin\theta$.

   $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt5}$

   (positive, as $\theta$ is in the first quadrant).

   Step 3 — Apply the half-angle identity.

   From $\cos\theta = 1 - 2\sin^2\frac\theta2$ and $\sin\theta = 2\sin\frac\theta2\cos\frac\theta2$, dividing gives the standard result

   $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$

   Step 4 — Substitute.

   $\tan\frac{\theta}{2} = \frac{1 - \dfrac{2}{\sqrt5}}{\dfrac{1}{\sqrt5}} = \left(1 - \frac{2}{\sqrt5}\right)\sqrt5 = \sqrt5 - 2$ …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.

   [!FORMULA] $\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} \text { where } \theta \in\left[-\frac{\pi}{8}, \frac{\pi}{8}\right] \text { is equal to }$

   (A) $\sin 2 \theta$ (B) $2 \cos \theta$ (C) $\cos 2 \theta$ (D) $2 \sin \theta$

   ›Reveal solutionSolution

   The expression simplifies by repeatedly applying the half-angle identity for cosine, and the given interval ensures all square roots are positive. The final result is $2\cos\theta$.

   We start with the nested radical

   $\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta}}}$

   and the condition $\theta \in \left[-\frac{\pi}{8},\frac{\pi}{8}\right]$.

   The key idea: each time we see $2+2\cos(\text{something})$, we can rewrite it using the identity

   $2+2\cos\alpha = 4\cos^2\frac{\alpha}{2},$

   so that the square root becomes $2\left|\cos\frac{\alpha}{2}\right|$. The interval for $\theta$ guarantees that every angle we encounter lies in a range where cosine is non‑negative, so we can drop the absolute value.

   ---

   1. Start from the innermost expression

   $2+2\cos 8\theta = 4\cos^2 4\theta.$

   Hence

   $\sqrt{2+2\cos 8\theta} = \sqrt{4\cos^2 4\theta} = 2|\cos 4\theta|.$

   Since $\theta\in[-\pi/8,\pi/8]$, we have $4\theta\in[-\pi/2,\pi/2]$, where cosine is non‑negative. So $|\cos 4\theta| = \cos 4\theta$, and the innermost radical becomes $2\cos 4\theta$.

   2. Move one level outward Now the expression is

   $\sqrt{2 + \bigl(2\cos 4\theta\bigr)} = \sqrt{2+2\cos 4\theta}.$

   Apply the same identity:

   $2+2\cos 4\theta = 4\cos^2 2\theta,$

   so

   $\sqrt{2+2\cos 4\theta} = 2|\cos 2\theta|.$

   Here $2\theta\in[-\pi/4,\pi/4]$, where cosine is positive. Thus $|\cos 2\theta| = \cos 2\theta$, giving $2\cos 2\theta$.

   3. Move to the outermost level We now have

   $\sqrt{2 + \bigl(2\cos 2\theta\bigr)} = \sqrt{2+2\cos 2\theta}.$

   Again,

   $2+2\cos 2\theta = 4\cos^2\theta,$

   so

   $\sqrt{2+2\cos 2\theta} = 2|\cos\theta|.$ …

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9. COMEDK 2025Set 2025-E1 markMCQ
   Q.$\sin ^{-1}(x-1)+\cos ^{-1}(x-3)+\tan ^{-1}\left(\frac{x}{2-x^2}\right)=\cos ^{-1} k+\pi$, then the value of ' $k$ ' is (A) $0$ (B) $-\frac{1}{\sqrt{2}}$ (C) $1$ (D) $\frac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   The key idea is to simplify the given inverse trigonometric sum by analyzing the domain and using known identities, leading to a single value for $k$. The correct option is (D).

   We start with the equation:

   $\sin^{-1}(x-1) + \cos^{-1}(x-3) + \tan^{-1}\left(\frac{x}{2-x^2}\right) = \cos^{-1} k + \pi.$

   ---

   1. Determine the domain of $x$.

   For $\sin^{-1}(x-1)$ to be defined, we need $-1 \le x-1 \le 1 \Rightarrow 0 \le x \le 2$.

   For $\cos^{-1}(x-3)$ to be defined, we need $-1 \le x-3 \le 1 \Rightarrow 2 \le x \le 4$.

   The intersection of these two intervals is $x = 2$ only. So the only possible value of $x$ is $2$.

   Watch out

   A common mistake is to forget that both inverse functions must be defined simultaneously. The intersection of their domains is a single point.

   ---

   2. Evaluate each term at $x = 2$.

   • $\sin^{-1}(2-1) = \sin^{-1}(1) = \frac{\pi}{2}$.
   • $\cos^{-1}(2-3) = \cos^{-1}(-1) = \pi$.
   • $\tan^{-1}\left(\frac{2}{2 - 4}\right) = \tan^{-1}\left(\frac{2}{-2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}$.

   ---

   3. Sum the left-hand side.

   $$$$

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10. COMEDK 2026Set 2026-A1 markMCQ
    Q.

    [!FORMULA] $\int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x=$

    (A) $\frac{\pi}{4} x-\frac{x^2}{4}+C$ (B) $\frac{\pi}{4} x-\frac{x^2}{2}+C$ (C) $\frac{\pi}{2} x-\frac{x^2}{4}+C$ (D) $\frac{\pi}{4}-\frac{x}{4}+C$

    ›Reveal solutionSolution

    The integrand simplifies to $\frac{\pi}{4} - \frac{x}{2}$ for $x$ in a suitable interval, so the integral is $\frac{\pi}{4}x - \frac{x^2}{4} + C$, matching option (A).

    ---

    The key insight is that the expression inside the arctangent can be dramatically simplified using trigonometric identities. The presence of $\sqrt{\frac{1-\sin x}{1+\sin x}}$ is a classic signal: it often equals $\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$ or a related form, depending on the quadrant. Once we recognize that, the arctangent and the tangent cancel, leaving a simple linear function of $x$. Then the integration is trivial.

    ---

    1. Simplify the radical using a half-angle identity. Recall that $1 - \sin x = (\sin(x/2) - \cos(x/2))^2$ and $1 + \sin x = (\sin(x/2) + \cos(x/2))^2$. For $x$ in a range where these are positive (e.g., $-\pi/2 < x < \pi/2$), we have:

    $\sqrt{\frac{1-\sin x}{1+\sin x}} = \frac{|\sin(x/2) - \cos(x/2)|}{|\sin(x/2) + \cos(x/2)|}.$

    Choosing a convenient interval (say $0 < x < \pi/2$) where both numerator and denominator are positive, we can drop the absolute values.

    2. Rewrite as a tangent of a difference. Divide numerator and denominator by $\cos(x/2)$ (assuming $\cos(x/2) \neq 0$):

    $\frac{\sin(x/2) - \cos(x/2)}{\sin(x/2) + \cos(x/2)} = \frac{\tan(x/2) - 1}{\tan(x/2) + 1}.$

    This is exactly $\tan\left(\frac{x}{2} - \frac{\pi}{4}\right)$ because

    $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B},$

    and with $A = x/2$, $B = \pi/4$, $\tan(\pi/4)=1$, we get:

    $\tan\left(\frac{x}{2} - \frac{\pi}{4}\right) = \frac{\tan(x/2) - 1}{1 + \tan(x/2)}.$

    Notice the denominator matches. So:

    $\sqrt{\frac{1-\sin x}{1+\sin x}} = \tan\left(\frac{x}{2} - \frac{\pi}{4}\right).$

    3. Apply the arctangent. Since $\tan^{-1}(\tan \theta) = \theta$ for $\theta$ in $(-\pi/2, \pi/2)$, we need $\frac{x}{2} - \frac{\pi}{4}$ to lie in that interval. For $0 < x < \pi/2$, this holds. Thus:

    $\tan^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) = \frac{x}{2} - \frac{\pi}{4}.$

    But note: $\frac{x}{2} - \frac{\pi}{4}$ is negative for small $x$. The arctangent of a negative number is negative, so this is fine. However, many textbooks prefer the positive form $\frac{\pi}{4} - \frac{x}{2}$ (since $\tan(\pi/4 - x/2) = \cot(\pi/4 + x/2)$ etc.). Let’s check:

    $\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan(x/2)}{1 + \tan(x/2)} = \frac{\tan(x/2) - 1}{\tan(x/2) + 1} \times (-1)?$

    Actually:

    $\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan(x/2)}{1 + \tan(x/2)} = -\frac{\tan(x/2) - 1}{\tan(x/2) + 1}.$

    That gives the negative of our expression. So the correct match is:

    $\sqrt{\frac{1-\sin x}{1+\sin x}} = \tan\left(\frac{x}{2} - \frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right).$

    Therefore:

    $\tan^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) = \frac{x}{2} - \frac{\pi}{4}.$

    Equivalently, we can write it as $-\left(\frac{\pi}{4} - \frac{x}{2}\right)$. For integration, the constant shift doesn’t matter; we’ll use $\frac{x}{2} - \frac{\pi}{4}$.

    4. Integrate.

    $\int \left( \frac{x}{2} - \frac{\pi}{4} \right) dx = \frac{x^2}{4} - \frac{\pi}{4}x + C.$

    But the answer choices have $\frac{\pi}{4}x - \frac{x^2}{4} + C$. That’s just the negative of our result. This suggests we might have chosen the opposite sign branch. Let’s re-evaluate: …

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11. COMEDK 2025Set 2025-E1 markMCQ
    Q.The value of $\tan \left\{\cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)-\frac{\pi}{2}\right\}$ is (A) $-1$ (B) $\frac{1}{\sqrt{2}}$ (C) $1$ (D) $-\frac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    The expression simplifies by recognizing that $\cos^{-1}(\sqrt{2}/2) = \pi/4$, so the argument becomes $\pi/4 - \pi/2 = -\pi/4$, and $\tan(-\pi/4) = -1$. The correct option is (A).

    ---

    The key insight is to evaluate the inverse cosine first. Inverse trig functions return an angle; once we know that angle, the whole expression becomes a simple tangent of a difference.

    1. Evaluate $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$. Recall that $\cos(\pi/4) = \sqrt{2}/2$ and the range of $\cos^{-1}$ is $[0, \pi]$. Since $\pi/4$ lies in that range, we have

    $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}.$

    2. Substitute into the original expression. The argument of the tangent becomes

    $\frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4}.$

    3. Compute $\tan(-\pi/4)$. Since $\tan$ is an odd function, $\tan(-\theta) = -\tan\theta$. And $\tan(\pi/4) = 1$. Therefore $\tan\left(-\frac{\pi}{4}\right) = -1.$ …

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12. COMEDK 2024Set 2024-E1 markMCQ
    Q.

    [!FORMULA] $\text { Value of } \cos 105^{\circ} \text { is }$

    (A) $\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$ (B) $-\frac{(\sqrt{3}-1)}{2 \sqrt{2}}$ (C) $-\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$ (D) $-\frac{(1-\sqrt{3})}{2 \sqrt{2}}$

    ›Reveal solutionSolution

    We use the cosine addition formula to express $\cos 105^\circ$ as $\cos(60^\circ+45^\circ)$, then evaluate exactly. The result is $-\frac{\sqrt{3}-1}{2\sqrt{2}}$, which corresponds to option (B).

    ---

    The key idea is that $105^\circ$ is not a standard angle on the unit circle, but it can be written as the sum of two familiar angles: $60^\circ$ and $45^\circ$. The cosine addition formula then lets us compute the exact value without a calculator.

    Why this works:

    The cosine addition formula, $\cos(A+B) = \cos A \cos B - \sin A \sin B$, is derived from the geometry of rotating points on the unit circle. It turns a messy angle into a combination of exact values we already know.

    ---

    1. Rewrite the angle

       $105^\circ = 60^\circ + 45^\circ$. Both $60^\circ$ and $45^\circ$ have known sine and cosine values.

    2. Apply the cosine addition formula

    $\cos(60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ$

    3. Substitute the exact values

    $\cos 60^\circ = \frac12,\quad \cos 45^\circ = \frac{\sqrt{2}}{2},\quad \sin 60^\circ = \frac{\sqrt{3}}{2},\quad \sin 45^\circ = \frac{\sqrt{2}}{2}$

    So:

    $\cos 105^\circ = \left(\frac12\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)$

    4. Simplify the expression Factor $\frac{\sqrt{2}}{4}$ out of both terms:

    $\cos 105^\circ = \frac{\sqrt{2}}{4} (1 - \sqrt{3})$

    5. Rationalize the denominator (optional but matches the options) Multiply numerator and denominator by $\sqrt{2}$: $$$$

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