Q.If $\cos(\theta + \phi) = m\cos(\theta - \phi)$, then prove that $\tan\theta = \dfrac{1 - m}{1 + m}\cot\phi$.

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

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

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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: …

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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: $$$$

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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

   $$$$

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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: …

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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 $\cos(\sin^{-1}y)$ …

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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:

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

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

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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: …

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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 …

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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: $\sin^2 x (\sin x + 1)^2$ …

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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): $\frac{3\pi}{4} - \alpha$ …

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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:

    $$$$

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