Q.Find the principal value of the following: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}$, $x \neq 0$

Showing the 12 most recent of 15 on this concept.

1. 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:

   For $x$ in $(0, \pi/2)$, $\sin x$ is positive, so $1-\sin x < 1+\sin x$, hence the fraction is less than 1, so its square root is less than 1. The arctangent of a number less than 1 is between $0$ and $\pi/4$. But $\frac{x}{2} - \frac{\pi}{4}$ is negative for $x<\pi/2$. So that can’t be right — the arctangent must be positive.

   The error: we should have taken the absolute value differently. In fact, for $0<x<\pi/2$, $\sin(x/2) < \cos(x/2)$, so $\sin(x/2) - \cos(x/2)$ is negative. The square root of the square is the absolute value, so:

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

   Now $\frac{\pi}{4} - \frac{x}{2}$ is positive for $x<\pi/2$, so the arctangent gives:

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

   That’s the correct branch.

   5. Integrate the corrected expression.

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

   This matches option (A).

   Watch out

   A common mistake is forgetting that $\sqrt{(\sin(x/2)-\cos(x/2))^2} = |\sin(x/2)-\cos(x/2)|$, not the raw difference. This flips the sign and leads to the wrong constant in the integral.

   Tip

   Always test a specific value (e.g., $x=0$) to check the sign: at $x=0$, the integrand is $\tan^{-1}(1) = \pi/4$, so the antiderivative should give $\pi/4 \cdot 0 - 0 + C = C$, consistent with $\frac{\pi}{4}x - \frac{x^2}{4} + C$ at $x=0$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

2. 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

   $\tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2},$

   provided $\frac{\theta}{2}$ lies in the principal range of $\tan^{-1}$, i.e., $(-\pi/2, \pi/2)$. Since $\theta = \tan^{-1}x$ is in $(-\pi/2, \pi/2)$, $\theta/2$ is in $(-\pi/4, \pi/4)$, which is safely inside that range.

   5. Back-substitute Since $\theta = \tan^{-1}x$, we get

   $\frac{\theta}{2} = \frac{1}{2}\tan^{-1}x.$

   Thus the simplest form is $\frac{1}{2}\tan^{-1}x$.

   Tip

   The identity $\frac{1-\cos\theta}{\sin\theta} = \tan\frac{\theta}{2}$ is the key shortcut — it avoids messing with double-angle formulas.

   Watch out

   A common mistake is to forget the domain: if $x$ is negative, $\theta$ is negative, but $\theta/2$ still lies in $(-\pi/4, \pi/4)$, so the simplification holds for all $x \neq 0$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.The derivative of $y=\sin ^2\left[\cot ^{-1}\left(\sqrt{\frac{\mathbf{1}-\boldsymbol{x}}{\mathbf{1}+\boldsymbol{x}}}\right)\right]$ is (A) $\frac{1}{2}$ (B) $\frac{1-x}{2}$ (C) $\frac{x}{2}$ (D) 0

   ›Reveal solutionSolution

   The expression simplifies to $y = \dfrac{1+x}{2}$, whose derivative is the constant $\dfrac{1}{2}$ — option (A).

   Concept

   Rather than differentiate the nested $\sin^2[\cot^{-1}(\cdot)]$ directly, simplify it first. Writing the inner angle as $\theta$ and using $\csc^2\theta = 1 + \cot^2\theta$ collapses $\sin^2\theta$ into a simple rational function of $x$, which is then trivial to differentiate.

   Solution

   1. Substitute: let $\theta = \cot^{-1}\!\left(\sqrt{\dfrac{1-x}{1+x}}\right)$, so $\cot\theta = \sqrt{\dfrac{1-x}{1+x}}$ and $\cot^2\theta = \dfrac{1-x}{1+x}$.
   2. Use the identity:

   $\csc^2\theta = 1 + \cot^2\theta = 1 + \frac{1-x}{1+x} = \frac{2}{1+x}.$

   3. Invert for $\sin^2\theta$:

   $\sin^2\theta = \frac{1}{\csc^2\theta} = \frac{1+x}{2}.$

   4. So $y = \dfrac{1+x}{2}$, and

   $\frac{dy}{dx} = \frac{1}{2}.$

   A quick check at $x = 0$: $\cot^{-1}(1) = \pi/4$ and $\sin^2(\pi/4) = \tfrac12 = \tfrac{1+0}{2}$, consistent with $y = \tfrac{1+x}{2}$.

   Watch out

   The expression is not constant, so the derivative is not zero; $\sin^2\theta = \dfrac{1+x}{2}$ genuinely depends on $x$, giving a slope of $\tfrac12$.

   ✓Final answer

   The correct option is (A), $\dfrac{1}{2}$.

   ANSWER: A

4. KCET 2026Set UNKNOWN1 markMCQ
   Q.If $\alpha$ and $\beta$ are acute angles such that $\alpha+\beta$ and $\alpha-\beta$ satisfy the equation $\tan^2\theta - 4\tan\theta + 1 = 0$, then $\alpha$ and $\beta$ are respectively (A) $45^{\circ}, 30^{\circ}$ (B) $75^{\circ}, 15^{\circ}$ (C) $30^{\circ}, 60^{\circ}$ (D) $60^{\circ}, 45^{\circ}$

   ›Reveal solutionSolution

   Since $\alpha+\beta$ and $\alpha-\beta$ are the two roots of the given quadratic in $\tan\theta$, solve the quadratic to identify the two angles, then recover $\alpha$ and $\beta$.

   Step 1 — Solve the quadratic for its roots

   $\tan^2\theta - 4\tan\theta + 1 = 0.$

   Using the quadratic formula with $t=\tan\theta$:

   $t = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}.$

   So the roots are $t_1 = 2+\sqrt{3}$ and $t_2 = 2-\sqrt{3}$.

   Step 2 — Recognize these as standard tangent values

   Recall the known values $\tan 75^\circ = 2+\sqrt{3}$ and $\tan 15^\circ = 2-\sqrt{3}$. Since $\tan(\alpha+\beta)$ and $\tan(\alpha-\beta)$ are precisely the two roots of this equation, and $\alpha,\beta$ are acute with $\alpha+\beta > \alpha-\beta$ (as $\beta>0$), we must have

   $\alpha+\beta = 75^\circ, \qquad \alpha-\beta = 15^\circ.$

   Step 3 — Solve for $\alpha$ and $\beta$

   Adding the two equations:

   $2\alpha = 90^\circ \implies \alpha = 45^\circ.$

   Subtracting:

   $2\beta = 60^\circ \implies \beta = 30^\circ.$

   ✓Final answer

   The correct option is (A) — $\alpha=45^{\circ}$, $\beta=30^{\circ}$.

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

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

   Therefore on $\left(0,\frac{\pi}{2}\right)$ the inverse cotangent reduces to a linear function with

   $u=\cot^{-1}\!\left(\text{arg}\right)=x-\tfrac{\pi}{6},\qquad \frac{du}{dx}=1.$

   Given $2y=u^2$, i.e. $y=\tfrac12 u^2$:

   $\frac{dy}{dx}=u\cdot\frac{du}{dx}=\left(x-\tfrac{\pi}{6}\right)\cdot 1=x-\tfrac{\pi}{6}.$

   ✓Final answer

   $\dfrac{dy}{dx}=x-\dfrac{\pi}{6}$, which is option (A).

6. 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.

   $\frac{\pi}{2} + \pi - \frac{\pi}{4} = \frac{2\pi}{4} + \frac{4\pi}{4} - \frac{\pi}{4} = \frac{5\pi}{4}.$

   So the equation becomes:

   $\frac{5\pi}{4} = \cos^{-1} k + \pi.$

   ---

   4. Solve for $\cos^{-1} k$.

   $\cos^{-1} k = \frac{5\pi}{4} - \pi = \frac{\pi}{4}.$

   Thus,

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

   ---

   Tip

   The domain restriction forced $x = 2$, turning a messy equation into a simple arithmetic check. Always check domains first in inverse trig problems.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. 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

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

   Watch out

   A common mistake is to forget that $\cos^{-1}(\sqrt{2}/2)$ could also be $-\pi/4$ if one thinks of cosine's evenness, but the principal value of $\cos^{-1}$ is always in $[0,\pi]$, so only $\pi/4$ is correct.

   Tip

   Whenever you see $\cos^{-1}(\sqrt{2}/2)$ or $\sin^{-1}(\sqrt{2}/2)$, immediately think of the standard angle $\pi/4$ (or $45^\circ$). This shortcut saves time and avoids algebraic clutter.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. 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

   $\frac{\sqrt{2}\cos 36^\circ}{\sqrt{2}\sin 36^\circ} = \frac{\cos 36^\circ}{\sin 36^\circ} = \cot 36^\circ.$

   4. Convert cotangent to tangent Since $\cot \theta = \tan(90^\circ - \theta)$, we have

   $\cot 36^\circ = \tan(90^\circ - 36^\circ) = \tan 54^\circ.$

   Tip

   A faster path: divide numerator and denominator by $\cos 9^\circ$ to get

   $\frac{1 + \tan 9^\circ}{1 - \tan 9^\circ} = \tan(45^\circ + 9^\circ) = \tan 54^\circ,$

   using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ with $A=45^\circ$, $B=9^\circ$. This is the cleanest method.

   Watch out

   A common mistake is to think $\frac{\cos + \sin}{\cos - \sin}$ simplifies to $\tan(\theta)$ with $\theta = 9^\circ$ or $18^\circ$, but the $45^\circ$ shift is essential.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. 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:

   $\frac{\sqrt{6}+\sqrt{2}}{4} - \frac{\sqrt{6}-\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}.$

   6. Put it all together

   $\frac{2 \cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}} = \frac{\sqrt{2}}{\frac{1}{2}} = 2\sqrt{2}.$

   Tip

   A faster path: after step 3, note that $\cos x - \sin x = \sqrt{2}\sin\left(\frac{\pi}{4} - x\right)$. For $x = \frac{\pi}{12}$, $\frac{\pi}{4} - \frac{\pi}{12} = \frac{\pi}{6}$, so $\cos x - \sin x = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}$. Then the same final step gives $2\sqrt{2}$.

   Watch out

   A common mistake is to forget that $\tan x + \cot x = \frac{1}{\sin x \cos x}$ only when both are defined; here at $x = \frac{\pi}{12}$ they are fine. Another pitfall is mis-evaluating $\sin\frac{\pi}{6}$ as $\frac{\sqrt{3}}{2}$ instead of $\frac{1}{2}$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

10. 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

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

    Since $\theta\in[-\pi/8,\pi/8]$, cosine is positive, so $|\cos\theta| = \cos\theta$. The entire expression equals $2\cos\theta$.

    ---

    Watch out

    A common mistake is to forget the absolute value when taking square roots of squares. The given interval is crucial: it ensures every cosine we encounter is non‑negative, so we can safely drop the absolute value. Without the interval, the answer could be $2|\cos\theta|$ instead.

    Tip

    This problem is a classic “nested radical” that unravels like an onion: each layer uses the same half‑angle identity. Recognizing the pattern saves time — you don’t need to compute each square root separately; just note that after $n$ steps you get $2\cos(8\theta/2^n)$.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. 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 $\sqrt{2}$:

    $\frac{\sqrt{2}}{4} (1 - \sqrt{3}) = \frac{2}{4\sqrt{2}} (1 - \sqrt{3}) = \frac{1 - \sqrt{3}}{2\sqrt{2}}$

    Since $1 - \sqrt{3}$ is negative, we can also write:

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

    Tip

    Notice that $1 - \sqrt{3} = -(\sqrt{3} - 1)$. This simple sign swap is the only difference between options (B) and (D). Option (D) is $-\frac{(1-\sqrt{3})}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$, which is positive — but $\cos 105^\circ$ is negative because $105^\circ$ lies in Quadrant II. So (D) is wrong.

    Watch out

    A common mistake is to forget that $\cos(105^\circ)$ is negative. Always check the quadrant: $90^\circ < 105^\circ < 180^\circ$ means cosine is negative. That eliminates options (A) and (D) immediately.

    ---

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. 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

       $\cot\frac{\pi}{2} = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0$.

    4. Apply the outer arcsine

       $\sin^{-1}(0) = 0$, since $\sin 0 = 0$ and $0$ lies in the principal range $[-\pi/2, \pi/2]$ of arcsine.

    Watch out

    A common mistake is to forget that $\cot(\pi/2)$ is $0$, not undefined. Also, be careful with $\frac{1}{2}\tan^{-1}\frac{1}{\sqrt{3}}$: it’s half of $\pi/6$, not $\tan^{-1}(1/(2\sqrt{3}))$.

    Tip

    The half-angle formula for tangent wasn’t even needed here because the half-angle term combined so nicely with the other angles to give exactly $\pi/2$. Always check if the sum simplifies before diving into complicated trig identities.

    ✓Final answer

    The correct option is (D).

    ANSWER: D