Q.Prove that $2\sin^2\frac{\pi}{6} + \csc^2\frac{7\pi}{6}\cos^2\frac{\pi}{3} = \frac{3}{2}$.

Showing the 12 most recent of 27 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Suppose 'a' and 'b' are non-zero constants satisfying the following system of equations $\boldsymbol{a} \sin ^3 x+\boldsymbol{b} \cos ^3 x=\sin x \cos x$ and $\mathbf{a} \sin x-\boldsymbol{b} \cos x=0$, then $\mathbf{2}\left(\boldsymbol{a}^6+\boldsymbol{b}^6\right)-\mathbf{3}\left(\boldsymbol{a}^4+\boldsymbol{b}^4\right)+\mathbf{1}=$ (A) 1 (B) -1 (C) 0 (D) $2 \sin ^2 x$

   ›Reveal solutionSolution

   The key idea is to use the second equation to relate $a$ and $b$ to $\tan x$, then substitute into the first equation to find a simple relation between $a$ and $b$. This reduces the expression $2(a^6+b^6)-3(a^4+b^4)+1$ to a constant independent of $x$, which evaluates to $0$.

   We start with two equations in $a$ and $b$ that also involve $x$:

   $a \sin^3 x + b \cos^3 x = \sin x \cos x \qquad(1)$

   $a \sin x - b \cos x = 0 \qquad(2)$

   The second equation is simpler: it gives a direct proportionality between $a$ and $b$. That’s our entry point.

   ---

   1. Use equation (2) to express $a$ in terms of $b$ (or vice versa). From $a \sin x = b \cos x$, we get

   $a = b \frac{\cos x}{\sin x} = b \cot x.$

   Equivalently, $b = a \tan x$. This relation will let us eliminate one variable.

   2. Substitute into equation (1). Replace $a$ with $b \cot x$ in (1):

   $(b \cot x) \sin^3 x + b \cos^3 x = \sin x \cos x.$

   Since $\cot x \sin^3 x = \frac{\cos x}{\sin x} \cdot \sin^3 x = \cos x \sin^2 x$, the left side becomes

   $b \cos x \sin^2 x + b \cos^3 x = b \cos x (\sin^2 x + \cos^2 x) = b \cos x.$

   So equation (1) simplifies beautifully to

   $b \cos x = \sin x \cos x.$

   3. Solve for $b$ (and then $a$). Assuming $\cos x \neq 0$ (if $\cos x = 0$, then from (2) we’d have $a \sin x = 0$ with $\sin x = \pm 1$, forcing $a=0$, contradicting non-zero constants), we can divide by $\cos x$:

   $b = \sin x.$

   Then from $a = b \cot x = \sin x \cdot \frac{\cos x}{\sin x} = \cos x$.

   So we have the elegant pair:

   $a = \cos x, \quad b = \sin x.$

   4. Now evaluate the required expression. We need

   $2(a^6 + b^6) - 3(a^4 + b^4) + 1.$

   Substitute $a = \cos x$, $b = \sin x$:

   $a^6 + b^6 = \cos^6 x + \sin^6 x, \quad a^4 + b^4 = \cos^4 x + \sin^4 x.$

   5. Simplify using trigonometric identities. Recall:

   $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x.$

   For the sixth powers, factor as sum of cubes:

   $\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x).$

   Since $\sin^2 x + \cos^2 x = 1$, this becomes

   $\sin^6 x + \cos^6 x = \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x.$

   Substitute $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$:

   $\sin^6 x + \cos^6 x = (1 - 2\sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x.$

   6. Plug into the expression.

   $2(a^6+b^6) - 3(a^4+b^4) + 1 = 2(1 - 3\sin^2 x \cos^2 x) - 3(1 - 2\sin^2 x \cos^2 x) + 1.$

   Expand:

   $= 2 - 6\sin^2 x \cos^2 x - 3 + 6\sin^2 x \cos^2 x + 1.$

   The terms $-6\sin^2 x \cos^2 x$ and $+6\sin^2 x \cos^2 x$ cancel. Then

   $2 - 3 + 1 = 0.$

   Watch out

   A common mistake is to forget that $a$ and $b$ are constants (not functions of $x$), but here the system forces them to equal $\cos x$ and $\sin x$ for some $x$. However, the final expression simplifies to a constant independent of $x$, so it’s valid.

   Tip

   The identity $\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$ is a neat shortcut; it’s worth remembering as a special case of $p^3+q^3 = (p+q)^3 - 3pq(p+q)$ with $p=\sin^2 x, q=\cos^2 x$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. COMEDK 2026Set 2026-A1 markMCQ
   Q.

   [!FORMULA] $\text { The expression } \frac{\tan \left(x-\frac{\pi}{2}\right) \cos \left(\frac{3 \pi}{2}+x\right)-\sin ^3\left(\frac{7 \pi}{2}-x\right)}{\cos \left(x-\frac{\pi}{2}\right) \tan \left(\frac{3 \pi}{2}+x\right)} \text { simplifies to: }$

   (A) $\sin ^2 x$ (B) $\cos ^2 x-\sin ^2 x$ (C) $1+\cos ^2 x$ (D) $-\left(1+\cos ^2 x\right)$

   ›Reveal solutionSolution

   Apply cofunction and shift identities to each term, then cancel. The expression simplifies to $\sin^2 x$ — option (A) — confirmed by a numeric check at $x=\pi/4$.

   Concept and intuition

   For angles like $x-\frac{\pi}{2}$ or $\frac{3\pi}{2}+x$, use the cofunction/shift identities (the unit-circle "ASTC" rule) to replace each trig function with a simpler one, possibly with a sign change. We simplify each factor, then combine.

   Solution

   Simplify each factor:

   • $\tan\left(x-\frac{\pi}{2}\right) = -\cot x = -\dfrac{\cos x}{\sin x}$
   • $\cos\left(\frac{3\pi}{2}+x\right) = \sin x$
   • $\sin\left(\frac{7\pi}{2}-x\right) = \sin\left(\frac{3\pi}{2}-x\right) = -\cos x$, so $\sin^3\left(\frac{7\pi}{2}-x\right) = -\cos^3 x$
   • $\cos\left(x-\frac{\pi}{2}\right) = \sin x$
   • $\tan\left(\frac{3\pi}{2}+x\right) = -\cot x = -\dfrac{\cos x}{\sin x}$

   Numerator:

   $\left(-\frac{\cos x}{\sin x}\right)(\sin x) - (-\cos^3 x) = -\cos x + \cos^3 x.$

   Denominator:

   $(\sin x)\left(-\frac{\cos x}{\sin x}\right) = -\cos x.$

   Ratio:

   $\frac{-\cos x + \cos^3 x}{-\cos x} = \frac{\cos x(\cos^2 x - 1)}{-\cos x} = \frac{\cos^2 x - 1}{-1} = 1 - \cos^2 x = \sin^2 x.$

   Numeric check at $x=\pi/4$: the expression evaluates to $\tfrac12$, and $\sin^2(\pi/4)=\tfrac12$ — consistent (whereas $-(1+\cos^2 x) = -1.5$, ruling out option D).

   Tip

   When you see $\frac{7\pi}{2}$, subtract $2\pi$ to bring it into a familiar range: $\frac{7\pi}{2} - 2\pi = \frac{3\pi}{2}$.

   ✓Final answer

   The expression simplifies to $\sin^2 x$ — option (A).

   ANSWER: A

3. COMEDK 2026Set 2026-A1 markMCQ
   Q.If $2 \sin \theta=\left(x+\frac{1}{x}\right)$, then $\sin 3 \theta+\frac{1}{2}\left(x^3+\frac{1}{x^3}\right)=$ (A) 1 (B) -1 (C) 3 (D) 0

   ›Reveal solutionSolution

   The key is to express $x + \frac{1}{x}$ in terms of $\sin\theta$, then use triple-angle identities to simplify $\sin 3\theta$ and $x^3 + \frac{1}{x^3}$; the expression simplifies to $0$, so the answer is (D).

   We start with the given relation:

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

   Our goal is to evaluate

   $\sin 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right).$

   ---

   Concept and Intuition

   The expression $x + \frac{1}{x}$ is symmetric and often appears with trigonometric substitutions. If we set $x = e^{i\theta}$ (or $x = \cos\theta + i\sin\theta$), then $x + \frac{1}{x} = 2\cos\theta$. But here we have $2\sin\theta$, so we need a shift: let $x = e^{i(\pi/2 - \theta)} = i e^{-i\theta}$ or simply use the identity $\sin\theta = \cos(\pi/2 - \theta)$. Alternatively, we can work algebraically:

   • $x^3 + \frac{1}{x^3}$ can be expressed in terms of $x + \frac{1}{x}$ using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$.
   • $\sin 3\theta$ expands to $3\sin\theta - 4\sin^3\theta$.

   Combining these will let us cancel terms.

   ---

   Step-by-step solution

   1. Express $x^3 + \frac{1}{x^3}$ in terms of $x + \frac{1}{x}$. Let $u = x + \frac{1}{x}$. Then

   $u^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) = x^3 + \frac{1}{x^3} + 3u.$

   Hence

   $x^3 + \frac{1}{x^3} = u^3 - 3u.$

   Here $u = 2\sin\theta$, so

   $x^3 + \frac{1}{x^3} = (2\sin\theta)^3 - 3(2\sin\theta) = 8\sin^3\theta - 6\sin\theta.$

   2. Compute $\frac{1}{2}\left(x^3 + \frac{1}{x^3}\right)$.

   $\frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = \frac{1}{2}(8\sin^3\theta - 6\sin\theta) = 4\sin^3\theta - 3\sin\theta.$

   3. Express $\sin 3\theta$ in terms of $\sin\theta$. Using the triple-angle identity:

   $\sin 3\theta = 3\sin\theta - 4\sin^3\theta.$

   4. Add the two parts.

   $\sin 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = (3\sin\theta - 4\sin^3\theta) + (4\sin^3\theta - 3\sin\theta).$

   The terms cancel exactly:

   $= 0.$

   ---

   Watch out

   A common mistake is to forget the factor $\frac{1}{2}$ or to misapply the expansion of $(x + 1/x)^3$. Always check that the cross term $3(x)(1/x)(x+1/x) = 3u$ is included.

   Tip

   Notice that $\sin 3\theta$ and $\frac{1}{2}(x^3 + 1/x^3)$ are opposites in terms of $\sin\theta$: one is $3s - 4s^3$, the other is $4s^3 - 3s$. This symmetry guarantees cancellation regardless of the value of $\theta$ (as long as $x$ is real and nonzero).

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. COMEDK 2026Set 2026-M1 markMCQ
   Q.If $\sin A+\sin 2 A=x$ and $\cos A+\cos 2 A=y$ then the value of the expression $\left(x^2+y^2\right)\left(x^2+y^2-3\right)$ equals (A) 0 (B) $3 y$ (C) $\frac{y}{2}$ (D) $2 y$

   ›Reveal solutionSolution

   The key idea is to express $x$ and $y$ in terms of $A$ using sum-to-product identities, then simplify $x^2+y^2$ to $2+2\cos A$. Substituting shows the given expression equals $2y$, so the correct option is (D).

   We start with

   $x = \sin A + \sin 2A, \quad y = \cos A + \cos 2A.$

   The expression we need is

   $(x^2 + y^2)(x^2 + y^2 - 3).$

   If we can find $x^2 + y^2$ in a simple form, the rest is just substitution.

   ---

   Why this approach works

   The sum of a sine and a cosine pair with different angles often simplifies using the sum-to-product formulas. Here, $\sin A + \sin 2A$ and $\cos A + \cos 2A$ are perfect candidates. Once we combine them, $x^2 + y^2$ becomes something like $2 + 2\cos A$, which is easy to handle.

   ---

   Step-by-step solution

   1. Apply sum-to-product identities

   $\sin A + \sin 2A = 2\sin\left(\frac{A+2A}{2}\right)\cos\left(\frac{A-2A}{2}\right) = 2\sin\frac{3A}{2}\cos\left(-\frac{A}{2}\right).$

   Since $\cos(-\theta) = \cos\theta$, we get

   $x = 2\sin\frac{3A}{2}\cos\frac{A}{2}.$

   Similarly,

   $\cos A + \cos 2A = 2\cos\left(\frac{A+2A}{2}\right)\cos\left(\frac{A-2A}{2}\right) = 2\cos\frac{3A}{2}\cos\frac{A}{2}.$

   So

   $y = 2\cos\frac{3A}{2}\cos\frac{A}{2}.$

   2. Compute $x^2 + y^2$

   $x^2 + y^2 = \left(2\sin\frac{3A}{2}\cos\frac{A}{2}\right)^2 + \left(2\cos\frac{3A}{2}\cos\frac{A}{2}\right)^2.$

   Factor out $4\cos^2\frac{A}{2}$:

   $x^2 + y^2 = 4\cos^2\frac{A}{2}\left(\sin^2\frac{3A}{2} + \cos^2\frac{3A}{2}\right).$

   The bracket is just $1$, so

   $x^2 + y^2 = 4\cos^2\frac{A}{2}.$

   3. Substitute into the target expression Let $t = x^2 + y^2 = 4\cos^2\frac{A}{2}$. Then

   $(x^2 + y^2)(x^2 + y^2 - 3) = t(t - 3) = 4\cos^2\frac{A}{2}\left(4\cos^2\frac{A}{2} - 3\right).$

   4. Simplify using a trigonometric identity Recall the triple-angle formula for cosine:

   $\cos 3\theta = 4\cos^3\theta - 3\cos\theta.$

   Here, set $\theta = \frac{A}{2}$. Then

   $\cos\frac{3A}{2} = 4\cos^3\frac{A}{2} - 3\cos\frac{A}{2}.$

   Multiply both sides by $\cos\frac{A}{2}$:

   $\cos\frac{3A}{2}\cos\frac{A}{2} = 4\cos^4\frac{A}{2} - 3\cos^2\frac{A}{2}.$

   But notice that $4\cos^4\frac{A}{2} - 3\cos^2\frac{A}{2} = \cos^2\frac{A}{2}\left(4\cos^2\frac{A}{2} - 3\right)$.

   So

   $\cos\frac{3A}{2}\cos\frac{A}{2} = \cos^2\frac{A}{2}\left(4\cos^2\frac{A}{2} - 3\right).$

   5. Relate back to our expression Our expression is $4\cos^2\frac{A}{2}\left(4\cos^2\frac{A}{2} - 3\right)$. That’s exactly $4$ times the right-hand side above:

   $4\cos^2\frac{A}{2}\left(4\cos^2\frac{A}{2} - 3\right) = 4 \cdot \cos\frac{3A}{2}\cos\frac{A}{2}.$

   But from step 1, $y = 2\cos\frac{3A}{2}\cos\frac{A}{2}$. So

   $4\cos\frac{3A}{2}\cos\frac{A}{2} = 2y.$

   Hence

   $(x^2 + y^2)(x^2 + y^2 - 3) = 2y.$

   6. Check the multiple-choice options The result is $2y$, which matches option (D).

   Watch out

   A common mistake is to forget that $x^2+y^2$ simplifies to $4\cos^2(A/2)$, not $4$. Always check the factor of $\cos^2(A/2)$ — it’s not always 1!

   Tip

   Spotting the triple-angle identity early saves time: once you have $x^2+y^2 = 4\cos^2(A/2)$, the expression $4\cos^2(A/2)(4\cos^2(A/2)-3)$ is exactly $2y$ because $y = 2\cos(3A/2)\cos(A/2)$ and the triple-angle formula links them.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\text { The expression } \frac{1-\tan ^2\left(\frac{\pi}{4}-A\right)}{1+\tan ^2\left(\frac{\pi}{4}-A\right)} \text { equals }$

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

   ›Reveal solutionSolution

   The expression simplifies to $\sin 2A$ by recognizing the tangent half‑angle form of sine and using the complementary angle identity. The correct option is (B).

   We start with the given expression:

   $\frac{1-\tan^2\left(\frac{\pi}{4}-A\right)}{1+\tan^2\left(\frac{\pi}{4}-A\right)}.$

   Concept and intuition:

   This looks exactly like the formula for $\cos 2\theta$ in terms of $\tan\theta$:

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

   So if we set $\theta = \frac{\pi}{4} - A$, the expression becomes $\cos\left[2\left(\frac{\pi}{4}-A\right)\right] = \cos\left(\frac{\pi}{2} - 2A\right)$. And we know $\cos(\frac{\pi}{2} - x) = \sin x$. That gives $\sin 2A$ directly.

   Let’s go step by step.

   1. Recognize the identity The standard double‑angle identity for cosine in terms of tangent is:

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

   This holds for all $\theta$ where $\tan\theta$ is defined.

   2. Substitute $\theta = \frac{\pi}{4} - A$ Then:

   $\frac{1-\tan^2\left(\frac{\pi}{4}-A\right)}{1+\tan^2\left(\frac{\pi}{4}-A\right)} = \cos\left[2\left(\frac{\pi}{4}-A\right)\right].$

   3. Simplify the argument

   $2\left(\frac{\pi}{4}-A\right) = \frac{\pi}{2} - 2A.$

   So the expression equals $\cos\left(\frac{\pi}{2} - 2A\right)$.

   4. Use the complementary angle identity

   $\cos\left(\frac{\pi}{2} - x\right) = \sin x.$

   With $x = 2A$, we get:

   $\cos\left(\frac{\pi}{2} - 2A\right) = \sin 2A.$

   Tip

   A quick check: try $A = 0$. Then the original expression becomes $\frac{1-\tan^2(\pi/4)}{1+\tan^2(\pi/4)} = \frac{1-1}{1+1}=0$, and $\sin 2A = \sin 0 = 0$. For $A = \pi/4$, the expression becomes $\frac{1-\tan^2(0)}{1+\tan^2(0)} = \frac{1-0}{1+0}=1$, and $\sin(2\cdot\pi/4)=\sin(\pi/2)=1$. This confirms the result.

   Watch out

   A common mistake is to forget the factor of 2 inside the sine. The expression simplifies to $\sin 2A$, not $\sin A$. Always check with a simple angle to avoid that trap.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

6. KCET 2026Set UNKNOWN1 markMCQ
   Q.If $\sin^{-1} x + \sin^{-1} y = \pi/2$, then $x^2$ is equal to (A) $1 - y^2$ (B) $1 + y^2$ (C) $\sqrt{1 - y^2}$ (D) $\sqrt{1 + y^2}$

   ›Reveal solutionSolution

   Isolate $\sin^{-1}x$, convert it to a cosine of $\sin^{-1}y$ using the complementary-angle identity, then square.

   Step 1 — Isolate $\sin^{-1}x$

   $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2} \implies \sin^{-1}x = \frac{\pi}{2} - \sin^{-1}y$

   Step 2 — Take sine of both sides

   $x = \sin\left(\frac{\pi}{2} - \sin^{-1}y\right) = \cos\left(\sin^{-1}y\right)$

   using $\sin(\pi/2 - \theta) = \cos\theta$.

   Step 3 — Simplify $\cos(\sin^{-1}y)$

   If $\theta = \sin^{-1}y$, then $\sin\theta = y$, and since $\theta \in [0, \pi/2]$ here (both inverse-sine terms are non-negative for their sum to equal $\pi/2$), $\cos\theta = \sqrt{1-y^2} \geq 0$. So:

   $x = \sqrt{1 - y^2}$

   Step 4 — Square both sides

   $x^2 = 1 - y^2$

   ✓Final answer

   The correct option is (A) — $x^2 = 1 - y^2$.

7. COMEDK 2025Set 2025-A1 markMCQ
   Q.If $\cos A=\frac{3}{4}$, then $\left(32 \sin \frac{A}{2} \sin \frac{5 A}{2}\right)=$ (A) 7 (B) 16 (C) 11 (D) 8

   ›Reveal solutionSolution

   Use the product-to-sum identity to rewrite the expression in terms of cosines, then substitute the given $\cos A = \frac{3}{4}$ and simplify to get a numeric value. The result is 11.

   We are given $\cos A = \frac{3}{4}$ and need to evaluate $32 \sin \frac{A}{2} \sin \frac{5A}{2}$. The direct approach is to transform the product of sines into a sum of cosines, which lets us use the known cosine value.

   1. Apply the product-to-sum identity Recall: $\sin x \sin y = \frac{1}{2}[\cos(x-y) - \cos(x+y)]$. Here $x = \frac{A}{2}$ and $y = \frac{5A}{2}$, so

   $\sin\frac{A}{2} \sin\frac{5A}{2} = \frac{1}{2}\left[\cos\left(\frac{A}{2} - \frac{5A}{2}\right) - \cos\left(\frac{A}{2} + \frac{5A}{2}\right)\right].$

   2. Simplify the arguments

   $\frac{A}{2} - \frac{5A}{2} = -\frac{4A}{2} = -2A, \quad \frac{A}{2} + \frac{5A}{2} = \frac{6A}{2} = 3A.$

   Since cosine is even, $\cos(-2A) = \cos 2A$. Thus

   $\sin\frac{A}{2} \sin\frac{5A}{2} = \frac{1}{2}\left[\cos 2A - \cos 3A\right].$

   3. Multiply by 32

   $32 \sin\frac{A}{2} \sin\frac{5A}{2} = 32 \cdot \frac{1}{2} (\cos 2A - \cos 3A) = 16(\cos 2A - \cos 3A).$

   4. Express $\cos 2A$ and $\cos 3A$ in terms of $\cos A$ Use double-angle: $\cos 2A = 2\cos^2 A - 1$. Use triple-angle: $\cos 3A = 4\cos^3 A - 3\cos A$. Given $\cos A = \frac{3}{4}$, compute:

   $\cos 2A = 2\left(\frac{3}{4}\right)^2 - 1 = 2\cdot\frac{9}{16} - 1 = \frac{18}{16} - 1 = \frac{2}{16} = \frac{1}{8}.$

   $\cos 3A = 4\left(\frac{3}{4}\right)^3 - 3\left(\frac{3}{4}\right) = 4\cdot\frac{27}{64} - \frac{9}{4} = \frac{108}{64} - \frac{9}{4}.$

   Convert $\frac{9}{4}$ to sixteenths: $\frac{9}{4} = \frac{36}{16} = \frac{144}{64}$. So

   $\cos 3A = \frac{108}{64} - \frac{144}{64} = -\frac{36}{64} = -\frac{9}{16}.$

   5. Substitute into the expression

   $16(\cos 2A - \cos 3A) = 16\left(\frac{1}{8} - \left(-\frac{9}{16}\right)\right) = 16\left(\frac{1}{8} + \frac{9}{16}\right).$

   Write $\frac{1}{8}$ as $\frac{2}{16}$:

   $16\left(\frac{2}{16} + \frac{9}{16}\right) = 16 \cdot \frac{11}{16} = 11.$

   Tip

   A common mistake is forgetting the sign when using $\cos(-2A) = \cos 2A$ or misapplying the triple-angle formula. Always check that $\cos 3A$ is negative here because $A$ is acute (since $\cos A = 3/4$), so $3A$ is in the second quadrant.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. COMEDK 2025Set 2025-A1 markMCQ
   Q.Simplified expression of $1-\frac{\sin ^2 y}{1+\cos y}+\frac{1+\cos y}{\sin y}-\frac{\sin y}{1-\cos y}$ is : (A) $\sin y$ (B) $\cos y$ (C) 1 (D) 0

   ›Reveal solutionSolution

   The expression simplifies to $\cos y$ after combining fractions, using the Pythagorean identity, and cancelling common factors. The correct option is (B).

   We start with the expression:

   $1 - \frac{\sin^2 y}{1+\cos y} + \frac{1+\cos y}{\sin y} - \frac{\sin y}{1-\cos y}$

   The goal is to simplify it to one of the given options. The presence of $\sin^2 y$ and $1\pm\cos y$ suggests using the identity $\sin^2 y = 1 - \cos^2 y$, which often helps when denominators involve $1+\cos y$ or $1-\cos y$.

   ---

   1. Simplify the first fraction

   $\frac{\sin^2 y}{1+\cos y}$

   Using $\sin^2 y = 1 - \cos^2 y = (1-\cos y)(1+\cos y)$, we get:

   $\frac{(1-\cos y)(1+\cos y)}{1+\cos y} = 1 - \cos y$

   So the first two terms become:

   $1 - (1 - \cos y) = \cos y$

   Now the whole expression is:

   $\cos y + \frac{1+\cos y}{\sin y} - \frac{\sin y}{1-\cos y}$

   ---

   2. Combine the remaining two fractions

   We have:

   $\frac{1+\cos y}{\sin y} - \frac{\sin y}{1-\cos y}$

   Find a common denominator: $\sin y (1-\cos y)$.

   $= \frac{(1+\cos y)(1-\cos y) - \sin^2 y}{\sin y (1-\cos y)}$

   ---

   3. Simplify the numerator

   $(1+\cos y)(1-\cos y) = 1 - \cos^2 y = \sin^2 y$

   So numerator becomes:

   $\sin^2 y - \sin^2 y = 0$

   Thus the whole fraction is $0$.

   ---

   4. Final result

   The expression reduces to:

   $\cos y + 0 = \cos y$

   Tip

   The key insight was rewriting $\sin^2 y$ as $1-\cos^2 y$ to cancel the first denominator immediately. Many students try to combine all four terms at once, but simplifying stepwise avoids messy algebra.

   Watch out

   A common mistake is to forget that $1 - \frac{\sin^2 y}{1+\cos y}$ is not the same as $\frac{1+\cos y - \sin^2 y}{1+\cos y}$ — that would be incorrect because the $1$ is not over the denominator. Always treat the $1$ as $\frac{1}{1}$ and combine carefully.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. COMEDK 2025Set 2025-E1 markMCQ
   Q.The value of $\frac{\sin ^2 20^{\circ}+\cos ^4 20^{\circ}}{\sin ^4 20^{\circ}+\cos ^2 20^{\circ}}$ is : (A) 0 (B) 2 (C) 1 (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The expression simplifies to 1 by noticing a symmetry between numerator and denominator and using the identity $\sin^2\theta + \cos^2\theta = 1$. The correct option is (C).

   We are asked to evaluate

   $\frac{\sin^2 20^\circ + \cos^4 20^\circ}{\sin^4 20^\circ + \cos^2 20^\circ}.$

   At first glance, this looks messy — different powers of sine and cosine. But the structure hints at a clever symmetry: the numerator has $\sin^2$ and $\cos^4$, while the denominator has $\sin^4$ and $\cos^2$. If we swap sine and cosine, numerator and denominator swap roles. That suggests the whole fraction might equal 1.

   Let’s check this idea step by step.

   1. Use the Pythagorean identity Recall that for any angle $\theta$,

   $\sin^2\theta + \cos^2\theta = 1.$

   We can rewrite $\cos^4\theta$ as $(\cos^2\theta)^2$ and $\sin^4\theta$ as $(\sin^2\theta)^2$.

   2. Rewrite numerator and denominator Let $s = \sin^2 20^\circ$ and $c = \cos^2 20^\circ$. Then $s + c = 1$. The expression becomes

   $\frac{s + c^2}{s^2 + c}.$

   3. Replace $c$ with $1 - s$ (or $s$ with $1 - c$) Since $c = 1 - s$, we have $c^2 = (1 - s)^2 = 1 - 2s + s^2$. So numerator:

   $s + c^2 = s + (1 - 2s + s^2) = 1 - s + s^2.$

   Denominator:

   $s^2 + c = s^2 + (1 - s) = 1 - s + s^2.$

   They are identical!

   4. Conclusion Since numerator = denominator, the fraction equals 1 for any angle satisfying $s + c = 1$, which is always true. So the value is exactly 1, independent of the specific angle $20^\circ$.

   Tip

   The symmetry trick works here: if you replace $\sin$ with $\cos$ and vice versa, numerator and denominator swap. That often signals the ratio is 1 — try it before doing heavy algebra.

   Watch out

   A common mistake is to try to compute $\sin 20^\circ$ and $\cos 20^\circ$ numerically. That’s unnecessary and error-prone. The simplification is purely algebraic using the fundamental identity.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

10. COMEDK 2025Set 2025-E1 markMCQ
    Q.If $\sin x+\sin ^2 x=1$ then $\cos ^8 x+2 \cos ^6 x+\cos ^4 x$ is equal to : (A) 0 (B) 1 (C) $-1$ (D) 2

    ›Reveal solutionSolution

    The key is to rewrite the given trigonometric condition as a quadratic in $\sin x$, then express $\cos^2 x$ in terms of $\sin x$ using the Pythagorean identity. The expression simplifies to $1$, so the answer is (B).

    ---

    Concept and Intuition

    The problem gives $\sin x + \sin^2 x = 1$. This looks like a quadratic in $\sin x$, but more importantly, it lets us find a simple relationship between $\sin x$ and $\cos^2 x$.

    Recall the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$.

    If we rearrange the given equation, we get $\sin x = 1 - \sin^2 x = \cos^2 x$.

    That’s the golden link: $\cos^2 x = \sin x$.

    Once we have that, the whole expression $\cos^8 x + 2\cos^6 x + \cos^4 x$ becomes a polynomial in $\sin x$, which we can simplify using the original condition.

    ---

    Step-by-step solution

    1. Rewrite the given condition $\sin x + \sin^2 x = 1$ Subtract $\sin^2 x$ from both sides: $\sin x = 1 - \sin^2 x$ But $1 - \sin^2 x = \cos^2 x$ (Pythagorean identity). So we have:

    $\cos^2 x = \sin x$

    2. Express higher powers of $\cos x$ in terms of $\sin x$

       • $\cos^4 x = (\cos^2 x)^2 = (\sin x)^2 = \sin^2 x$
       • $\cos^6 x = (\cos^2 x)^3 = (\sin x)^3 = \sin^3 x$
       • $\cos^8 x = (\cos^2 x)^4 = (\sin x)^4 = \sin^4 x$

    3. Substitute into the target expression

       The expression becomes:

    $\cos^8 x + 2\cos^6 x + \cos^4 x = \sin^4 x + 2\sin^3 x + \sin^2 x$

    4. Factor the polynomial in $\sin x$ Notice that $\sin^4 x + 2\sin^3 x + \sin^2 x = \sin^2 x (\sin^2 x + 2\sin x + 1)$ The quadratic factor is a perfect square: $\sin^2 x + 2\sin x + 1 = (\sin x + 1)^2$ So we have:

    $\sin^2 x (\sin x + 1)^2$

    5. Use the original condition to simplify further

       From $\sin x + \sin^2 x = 1$, we can write $\sin x (1 + \sin x) = 1$

       That is: $\sin x (\sin x + 1) = 1$

       Square both sides: $\sin^2 x (\sin x + 1)^2 = 1^2 = 1$

    6. Conclusion

       Therefore, the entire expression equals $1$.

    ---

    Watch out

    A common mistake is to solve $\sin x + \sin^2 x = 1$ as a quadratic and get $\sin x = \frac{-1 \pm \sqrt{5}}{2}$, then try to compute $\cos^2 x$ from that. That works but is messy. The elegant shortcut is to notice $\cos^2 x = \sin x$ directly from the rearrangement.

    Tip

    Whenever you see $\sin x + \sin^2 x = 1$, immediately think $\cos^2 x = \sin x$. It turns a trigonometric expression into a simple algebraic one.

    ---

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. COMEDK 2025Set 2025-E1 markMCQ
    Q.If $\tan \alpha=\frac{1}{7}$ and $\sin \beta=\frac{1}{\sqrt{10}}, \quad 0<\alpha, \beta<\frac{\pi}{2}$ then $2 \beta$ is equal to (A) $\frac{\pi}{8}-\alpha$ (B) $\frac{\pi}{4}-\alpha$ (C) $\frac{3 \pi}{8}-\frac{\alpha}{2}$ (D) $\frac{3 \pi}{4}-\alpha$

    ›Reveal solutionSolution

    Using the given $\tan\alpha$ and $\sin\beta$, we compute $\tan(2\beta)$ and compare it with $\tan(\frac{\pi}{4}-\alpha)$ to find that $2\beta = \frac{\pi}{4} - \alpha$, so the answer is (B).

    We are told $\tan\alpha = \frac{1}{7}$ and $\sin\beta = \frac{1}{\sqrt{10}}$, with both angles in the first quadrant ($0<\alpha,\beta<\frac{\pi}{2}$). The question asks which expression equals $2\beta$.

    The natural idea: find $\tan(2\beta)$ directly from $\sin\beta$, then see which of the given options has the same tangent. Since all options involve $\alpha$ and constants like $\frac{\pi}{4}$, we can compute $\tan(\text{option})$ and match.

    Step-by-step reasoning

    1. Find $\cos\beta$ and $\tan\beta$ Given $\sin\beta = \frac{1}{\sqrt{10}}$ and $\beta$ in $(0,\pi/2)$, we have

    $\cos\beta = \sqrt{1-\sin^2\beta} = \sqrt{1-\frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}.$

    Hence

    $\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}.$

    2. Compute $\tan(2\beta)$ Using the double-angle formula:

    $\tan(2\beta) = \frac{2\tan\beta}{1-\tan^2\beta} = \frac{2\cdot\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2} = \frac{\frac{2}{3}}{1-\frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \cdot \frac{9}{8} = \frac{18}{24} = \frac{3}{4}.$

    So $\tan(2\beta) = \frac{3}{4}$.

    3. Check each option by taking tangent

       We know $\tan\alpha = \frac{1}{7}$. We'll compute $\tan(\text{option})$ and see which equals $\frac{3}{4}$.

       • Option (A): $\frac{\pi}{8} - \alpha$

         $\tan(\frac{\pi}{8})$ is not a standard simple value (it's $\sqrt{2}-1$), so this is unlikely to match $\frac{3}{4}$ with $\tan\alpha=1/7$. We can check later if needed, but let's first test the more promising ones.

       • Option (B): $\frac{\pi}{4} - \alpha$

         Use the tangent subtraction formula:

    $\tan\left(\frac{\pi}{4} - \alpha\right) = \frac{\tan\frac{\pi}{4} - \tan\alpha}{1 + \tan\frac{\pi}{4}\tan\alpha} = \frac{1 - \frac{1}{7}}{1 + 1\cdot\frac{1}{7}} = \frac{\frac{6}{7}}{\frac{8}{7}} = \frac{6}{8} = \frac{3}{4}.$

     This matches $\tan(2\beta)$ exactly.
    

    • Option (C): $\frac{3\pi}{8} - \frac{\alpha}{2}$

      $\tan(\frac{3\pi}{8})$ is $\sqrt{2}+1$, and with $\tan(\alpha/2)$ involved, it's messy and unlikely to simplify to $\frac{3}{4}$ given $\tan\alpha=1/7$. We can verify later if needed.

    • Option (D): $\frac{3\pi}{4} - \alpha$

      $\tan\left(\frac{3\pi}{4} - \alpha\right) = \tan\left(\pi - \frac{\pi}{4} - \alpha\right) = \tan\left(\frac{3\pi}{4} - \alpha\right)$. Since $\tan(\frac{3\pi}{4}) = -1$, we get

    $\tan\left(\frac{3\pi}{4} - \alpha\right) = \frac{-1 - \frac{1}{7}}{1 + (-1)\cdot\frac{1}{7}} = \frac{-\frac{8}{7}}{\frac{6}{7}} = -\frac{4}{3},$

     which is not $\frac{3}{4}$.
    

    4. Confirm that $2\beta$ and $\frac{\pi}{4}-\alpha$ are in the same quadrant

    Since $0<\alpha,\beta<\pi/2$, we have $0<2\beta<\pi$ and $0<\frac{\pi}{4}-\alpha<\frac{\pi}{4}$ (because $\alpha>0$). Both are acute angles (positive and less than $\pi/2$), so having equal tangents implies they are equal.

    Hence $2\beta = \frac{\pi}{4} - \alpha$.

    Tip

    The key trick: when two acute angles have the same tangent, they are equal. This avoids worrying about other possible angle shifts.

    Watch out

    A common mistake is to forget that $\tan(\theta) = \tan(\theta + \pi)$; but here both angles are in $(0,\pi/2)$, so no ambiguity.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. COMEDK 2025Set 2025-M1 markMCQ
    Q.If $\sin A+\sin B=-\frac{21}{65}, \cos A+\cos B=-\frac{27}{65}$ and $\pi<A-B<3 \pi$, then the value of $\cos \left(\frac{A-B}{2}\right)$ is (A) $\frac{3}{\sqrt{130}}$ (B) $\frac{6}{65}$ (C) $-\frac{6}{65}$ (D) $-\frac{3}{\sqrt{130}}$

    ›Reveal solutionSolution

    Using sum-to-product identities, we find $\cos\frac{A-B}{2}$ from the ratio of the given sums; the sign is determined by the quadrant condition $\pi<A-B<3\pi$, giving $-\frac{3}{\sqrt{130}}$.

    We are given:

    $\sin A + \sin B = -\frac{21}{65}, \quad \cos A + \cos B = -\frac{27}{65}, \quad \pi < A-B < 3\pi.$

    We need $\cos\left(\frac{A-B}{2}\right)$.

    Concept & Intuition

    When we have sums of sines and cosines of two angles, the natural tool is the sum-to-product formulas. They rewrite each sum as a product of a sine or cosine of the average and a sine or cosine of the half-difference. The ratio of the two given sums then isolates $\tan\frac{A-B}{2}$, from which we can find $\cos\frac{A-B}{2}$. The range condition on $A-B$ tells us the sign of the half-angle cosine.

    Step-by-step solution

    1. Apply sum-to-product identities

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

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

    So the given equations become:

    $2 \sin\frac{A+B}{2} \cos\frac{A-B}{2} = -\frac{21}{65}$

    $2 \cos\frac{A+B}{2} \cos\frac{A-B}{2} = -\frac{27}{65}$

    2. Divide the two equations (provided $\cos\frac{A-B}{2} \neq 0$, which we’ll verify later)

    $\frac{2 \sin\frac{A+B}{2} \cos\frac{A-B}{2}}{2 \cos\frac{A+B}{2} \cos\frac{A-B}{2}} = \frac{-\frac{21}{65}}{-\frac{27}{65}}$

    The factors $2$ and $\cos\frac{A-B}{2}$ cancel, and the negatives cancel, giving:

    $\tan\frac{A+B}{2} = \frac{21}{27} = \frac{7}{9}$

    3. Find $\cos\frac{A-B}{2}$ using one of the original equations From the cosine sum equation:

    $2 \cos\frac{A+B}{2} \cos\frac{A-B}{2} = -\frac{27}{65}$

    We know $\tan\frac{A+B}{2} = \frac{7}{9}$, so we can find $\cos\frac{A+B}{2}$.

    Since $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$, we can think of a right triangle with opposite $7$ and adjacent $9$; the hypotenuse is $\sqrt{7^2+9^2} = \sqrt{49+81} = \sqrt{130}$.

    Hence:

    $\cos\frac{A+B}{2} = \pm \frac{9}{\sqrt{130}}$

    The sign depends on the quadrant of $\frac{A+B}{2}$, but we don’t need it directly — we can square to avoid sign ambiguity.

    Substitute into the equation:

    $2 \left(\pm \frac{9}{\sqrt{130}}\right) \cos\frac{A-B}{2} = -\frac{27}{65}$

    Multiply both sides by $\sqrt{130}$:

    $\pm 18 \cos\frac{A-B}{2} = -\frac{27\sqrt{130}}{65}$

    Simplify $-\frac{27\sqrt{130}}{65} = -\frac{27}{65}\sqrt{130}$. Notice $\frac{27}{65} = \frac{27}{65}$ and $18 = \frac{18}{1}$. Divide both sides by $\pm 18$:

    $\cos\frac{A-B}{2} = \mp \frac{27\sqrt{130}}{65 \cdot 18} = \mp \frac{27\sqrt{130}}{1170}$

    Simplify $\frac{27}{1170} = \frac{3}{130}$ (since $27 \div 9 = 3$, $1170 \div 9 = 130$). So:

    $\cos\frac{A-B}{2} = \mp \frac{3\sqrt{130}}{130} = \mp \frac{3}{\sqrt{130}}$

    That is, $\cos\frac{A-B}{2} = \pm \frac{3}{\sqrt{130}}$, but the sign is opposite to the sign of $\cos\frac{A+B}{2}$.

    4. Determine the correct sign using the given range We are told $\pi < A-B < 3\pi$. Divide by 2:

    $\frac{\pi}{2} < \frac{A-B}{2} < \frac{3\pi}{2}$

    In this interval, cosine is negative (since angles in quadrants II and III have negative cosine).

    Therefore $\cos\frac{A-B}{2}$ must be negative.

    So we choose the negative value:

    $\cos\frac{A-B}{2} = -\frac{3}{\sqrt{130}}$

    Watch out

    A common mistake is to forget that dividing the two equations cancels the sign of $\cos\frac{A-B}{2}$, so you must use the range condition separately to decide the sign. Also, note that $\cos\frac{A+B}{2}$ could be positive or negative, but the product with $\cos\frac{A-B}{2}$ is fixed; the sign of $\cos\frac{A-B}{2}$ is determined solely by the half-difference interval.

    Tip

    The ratio $\frac{\sin A+\sin B}{\cos A+\cos B} = \tan\frac{A+B}{2}$ is a neat shortcut — it avoids dealing with the factor $\cos\frac{A-B}{2}$ until the final step.

    ✓Final answer

    The correct option is (D).

    ANSWER: D