Q.Show that $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}=\tan^{-1}\frac{63}{16}$.

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.$\sin\left(\tan^{-1}\frac{4}{5} + \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{1}{9} - \tan^{-1}\frac{1}{7}\right) =$ (A) $\frac{1}{2}$ (B) $\frac{1}{\sqrt{2}}$ (C) $\frac{\sqrt{3}}{2}$ (D) $1$

   ›Reveal solutionSolution

   Pair the four inverse tangents so each pair collapses to $\tfrac{\pi}{4}$: $\tan^{-1}\tfrac45+\tan^{-1}\tfrac19=\tfrac{\pi}{4}$ and $\tan^{-1}\tfrac43-\tan^{-1}\tfrac17=\tfrac{\pi}{4}$. Their sum is $\tfrac{\pi}{2}$, so the sine is $1$.

   Group the terms cleverly using

   $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy},\qquad \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}.$

   Pair 1: $\tan^{-1}\dfrac45+\tan^{-1}\dfrac19$

   $\frac{\frac45+\frac19}{1-\frac45\cdot\frac19}=\frac{\frac{36+5}{45}}{\frac{45-4}{45}}=\frac{41/45}{41/45}=1\ \Rightarrow\ \tan^{-1}(1)=\frac{\pi}{4}.$

   Pair 2: $\tan^{-1}\dfrac43-\tan^{-1}\dfrac17$ …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If both $\alpha$ and $\beta$ lie in $\left(-\dfrac{\pi}{4}, 0\right)$, $\sin(\alpha+\beta) = -\dfrac{33}{65}$ and $\sin\alpha = -\dfrac{3}{5}$, then $\tan(2\alpha+\beta) =$ (A) $\dfrac{-9}{323}$ (B) $\dfrac{-9}{125}$ (C) $\dfrac{75}{323}$ (D) $\dfrac{-12}{5}$

   ›Reveal solutionSolution

   With $\cos\alpha=\tfrac45$ and $\cos(\alpha+\beta)=\tfrac{56}{65}$, writing $2\alpha+\beta=\alpha+(\alpha+\beta)$ gives $\tan(2\alpha+\beta)=-\dfrac{12}{5}$ — option (D).

   Both $\alpha$ and $\beta$ lie in $\left(-\tfrac{\pi}{4},0\right)$, so their sines are negative and their cosines positive.

   From $\sin\alpha=-\tfrac35$: $\cos\alpha=+\sqrt{1-\tfrac{9}{25}}=\tfrac45$, hence $\tan\alpha=-\tfrac34$.

   Since $\alpha+\beta\in\left(-\tfrac{\pi}{2},0\right)$, $\cos(\alpha+\beta)>0$. From $\sin(\alpha+\beta)=-\tfrac{33}{65}$:

   $\cos(\alpha+\beta)=\sqrt{1-\tfrac{1089}{4225}}=\sqrt{\tfrac{3136}{4225}}=\tfrac{56}{65},\qquad \tan(\alpha+\beta)=-\tfrac{33}{56}.$

   Write $2\alpha+\beta=\alpha+(\alpha+\beta)$ and apply the tangent addition formula: …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.

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

   (A) $\frac{1}{\cos x \sqrt{\cos 2x}} + \frac{2 \cos x}{3 - \cos 2x}$ (B) $\frac{\sec x}{\sqrt{1 + \tan^2 x}} + \frac{\cos x}{1 - \sin^2 x}$ (C) $\frac{\sec x \tan x}{\sqrt{\cos 2x}} + \frac{\sin x}{1 + \cos 2x}$ (D) $\frac{\sec x}{1 - \tan^2 x} + \frac{\cos x}{\sqrt{1 + \sin^2 x}}$

   ›Reveal solutionSolution

   The derivative simplifies to $\frac{\sec x}{\sqrt{\cos 2x}} + \frac{\cos x}{1+\sin^2 x}$, which matches option (A) after rewriting $\frac{2\cos x}{3-\cos 2x}$ as $\frac{\cos x}{1+\sin^2 x}$.

   We need to differentiate

   $f(x) = \sin^{-1}(\tan x) + \tan^{-1}(\sin x)$

   and match the result to one of the given options. The key is to differentiate each term carefully, simplify using trigonometric identities, and then compare forms.

   ---

   1. Differentiate $\sin^{-1}(\tan x)$

   Recall:

   $\frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$

   Here $u = \tan x$, so $\frac{du}{dx} = \sec^2 x$.

   Thus:

   $\frac{d}{dx} \sin^{-1}(\tan x) = \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}}$

   Now simplify the denominator:

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

   So

   $\sqrt{1 - \tan^2 x} = \frac{\sqrt{\cos 2x}}{|\cos x|}$

   Since we are likely in a domain where $\cos x > 0$ (to keep $\tan x$ within $[-1,1]$ for the inverse sine), we take $\cos x > 0$ and drop absolute value.

   Therefore:

   $\frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} = \frac{1/\cos^2 x}{\sqrt{\cos 2x}/\cos x} = \frac{1}{\cos x \sqrt{\cos 2x}}$

   So the first term is $\frac{1}{\cos x \sqrt{\cos 2x}}$.

   ---

   2. Differentiate $\tan^{-1}(\sin x)$

   Recall:

   $\frac{d}{dx} \tan^{-1} u = \frac{1}{1+u^2} \cdot \frac{du}{dx}$

   Here $u = \sin x$, so $\frac{du}{dx} = \cos x$.

   Thus:

   $\frac{d}{dx} \tan^{-1}(\sin x) = \frac{\cos x}{1 + \sin^2 x}$

   ---

   3. Combine and compare with options

   The derivative is:

   $\frac{1}{\cos x \sqrt{\cos 2x}} + \frac{\cos x}{1 + \sin^2 x}$

   Now check the options:

   • (A) $\frac{1}{\cos x \sqrt{\cos 2x}} + \frac{2 \cos x}{3 - \cos 2x}$ Note: $3 - \cos 2x = 3 - (1 - 2\sin^2 x) = 2 + 2\sin^2 x = 2(1 + \sin^2 x)$. …

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4. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $\cos \theta = -\dfrac{3}{5}$ and $\pi < \theta < \dfrac{3\pi}{2}$, then $\tan \dfrac{\theta}{2} + \sin \dfrac{\theta}{2} + 2\cos \dfrac{\theta}{2} =$ (A) $-1$ (B) $1$ (C) $-2$ (D) $2$

   ›Reveal solutionSolution

   With $\theta$ in the third quadrant, $\tfrac{\theta}{2}$ lies in the second quadrant: $\tan\tfrac{\theta}{2}=-2$, $\sin\tfrac{\theta}{2}=\tfrac{2}{\sqrt5}$, $\cos\tfrac{\theta}{2}=-\tfrac{1}{\sqrt5}$, giving the sum $-2$ — option (C).

   Locating the angles

   Given $\cos\theta=-\dfrac35$ with $\pi<\theta<\dfrac{3\pi}{2}$ (third quadrant), $\sin\theta<0$:

   $\sin\theta = -\sqrt{1-\tfrac{9}{25}} = -\frac45.$

   Since $\pi<\theta<\dfrac{3\pi}{2}$, we have $\dfrac{\pi}{2}<\dfrac{\theta}{2}<\dfrac{3\pi}{4}$ — the second quadrant, where $\sin\tfrac{\theta}{2}>0$ and $\cos\tfrac{\theta}{2}<0$.

   Half-angle values

   $\sin^2\frac{\theta}{2} = \frac{1-\cos\theta}{2} = \frac{1+\frac35}{2} = \frac{4}{5} \;\Rightarrow\; \sin\frac{\theta}{2} = \frac{2}{\sqrt5},$ …

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5. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $\left(2\sin^{-1}x\right)^3 = \pi^3 - \left(2\cos^{-1}x\right)^3$, then one of the values of $\cos\left(2\sin^{-1}x - 3\cos^{-1}x\right)$ is (A) $-1$ (B) $\dfrac{\pi}{2}$ (C) $1$ (D) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   The key idea is to treat the given equation as a sum of cubes and use the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ to solve for $\sin^{-1}x$ and $\cos^{-1}x$ individually. The required expression then simplifies to $\cos\left(-\frac{\pi}{2}\right) = 0$, but since $0$ is not among the options, we check the other possible branch and find the value is $1$.

   The problem gives a cubic relation between inverse trigonometric functions. The natural instinct is to expand or factor, but the presence of cubes and the known identity linking $\sin^{-1}x$ and $\cos^{-1}x$ points directly to a sum-of-cubes factorization.

   Let $a = 2\sin^{-1}x$ and $b = 2\cos^{-1}x$. Then the equation becomes $a^3 + b^3 = \pi^3$. Factor the sum of cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Now, $a+b = 2(\sin^{-1}x + \cos^{-1}x) = 2 \cdot \frac{\pi}{2} = \pi$. So we have $\pi (a^2 - ab + b^2) = \pi^3$, which simplifies to $a^2 - ab + b^2 = \pi^2$.

   Now we know $a+b = \pi$. Square both sides: $a^2 + 2ab + b^2 = \pi^2$. Subtract the earlier equation $a^2 - ab + b^2 = \pi^2$ from this: $(a^2 + 2ab + b^2) - (a^2 - ab + b^2) = \pi^2 - \pi^2$, giving $3ab = 0$, so $ab = 0$.

   Thus either $a=0$ or $b=0$.

   1. Case 1: $a = 0 \implies 2\sin^{-1}x = 0 \implies \sin^{-1}x = 0 \implies x = 0$. Then $\cos^{-1}x = \frac{\pi}{2}$, so $b = 2 \cdot \frac{\pi}{2} = \pi$. This satisfies $a+b = \pi$ and $ab=0$.

   2. Case 2: $b = 0 \implies 2\cos^{-1}x = 0 \implies \cos^{-1}x = 0 \implies x = 1$. Then $\sin^{-1}x = \frac{\pi}{2}$, so $a = 2 \cdot \frac{\pi}{2} = \pi$. This also satisfies the conditions.

   Now we need $\cos\left(2\sin^{-1}x - 3\cos^{-1}x\right)$. …

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6. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.$\cos \dfrac{7\pi}{8} + \cos \dfrac{\pi}{4} + \cos \left( -\dfrac{\pi}{8} \right) - 1 =$ (A) $4 \cos \dfrac{\pi}{16} \cos \dfrac{3\pi}{4} \cos \dfrac{5\pi}{8}$ (B) $4 \cos \dfrac{\pi}{16} \cos \dfrac{\pi}{8} \sin \dfrac{5\pi}{8}$ (C) $4 \cos \dfrac{\pi}{16} \cos \dfrac{3\pi}{8} \cos \dfrac{9\pi}{16}$ (D) $-4 \cos \dfrac{\pi}{16} \cos \dfrac{5\pi}{8} \cos \dfrac{\pi}{16}$

   ›Reveal solutionSolution

   The expression simplifies using sum-to-product identities and angle transformations into a product of cosines; the final result matches option (C).

   The key here is to notice that the given expression is a sum of three cosines minus 1. Direct evaluation of each cosine numerically would work, but the problem expects you to recognise a pattern that collapses into a product. The angles $\frac{7\pi}{8}$, $\frac{\pi}{4}$, and $-\frac{\pi}{8}$ are all related — they are symmetric about $\frac{\pi}{2}$ and $\pi$, which makes sum-to-product formulas the natural tool.

   1. Simplify the negative angle. Cosine is an even function: $\cos(-\theta) = \cos \theta$. So

   $\cos\left(-\frac{\pi}{8}\right) = \cos\frac{\pi}{8}.$

   The expression becomes

   $\cos\frac{7\pi}{8} + \cos\frac{\pi}{4} + \cos\frac{\pi}{8} - 1.$

   2. Group the first and last terms. $\cos\frac{7\pi}{8}$ and $\cos\frac{\pi}{8}$ are a natural pair because $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$. Using $\cos(\pi - \theta) = -\cos\theta$, we get

   $\cos\frac{7\pi}{8} = -\cos\frac{\pi}{8}.$

   So $\cos\frac{7\pi}{8} + \cos\frac{\pi}{8} = 0$. That cancels neatly.

   The expression reduces to

   $\cos\frac{\pi}{4} - 1.$

   3. Evaluate directly. $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so

   $\frac{\sqrt{2}}{2} - 1.$

   That is a simple number. But the answer choices are all products of four cosines (or sines) — so clearly this direct evaluation is not the intended path. Something is off.

   Watch out

   The cancellation in step 2 is correct, but it leads to a constant, not a product. That means the original expression must be interpreted differently — perhaps the $-1$ is part of a larger identity, or the sum is meant to be transformed before simplifying individual terms. Let’s re-approach without cancelling prematurely.

   4. Use sum-to-product on $\cos\frac{7\pi}{8} + \cos\frac{\pi}{8}$. The sum-to-product formula:

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

   Take $A = \frac{7\pi}{8}$, $B = \frac{\pi}{8}$. Then

   $\frac{A+B}{2} = \frac{\frac{7\pi}{8} + \frac{\pi}{8}}{2} = \frac{\pi}{2}, \quad \frac{A-B}{2} = \frac{\frac{7\pi}{8} - \frac{\pi}{8}}{2} = \frac{3\pi}{8}.$

   So

   $\cos\frac{7\pi}{8} + \cos\frac{\pi}{8} = 2\cos\frac{\pi}{2}\cos\frac{3\pi}{8} = 2 \cdot 0 \cdot \cos\frac{3\pi}{8} = 0.$

   That confirms the cancellation. So the expression is indeed $\cos\frac{\pi}{4} - 1$, which is a constant. But none of the options equal that constant for all angles — unless the options are meant to be identities that hold for any angle? No, the options are specific products.

   This suggests the problem might have a misprint, or the $-1$ is actually part of a different grouping. Let’s check if the intended expression is something like

   $\cos\frac{7\pi}{8} + \cos\frac{\pi}{4} + \cos\left(-\frac{\pi}{8}\right) - 1$

   but with the $-1$ meant to be combined with $\cos\frac{\pi}{4}$ using a half-angle identity.

   5. Try a different grouping: combine $\cos\frac{\pi}{4} - 1$ using a half-angle. Recall $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$, so $\cos\theta - 1 = -2\sin^2\frac{\theta}{2}$. With $\theta = \frac{\pi}{4}$,

   $\cos\frac{\pi}{4} - 1 = -2\sin^2\frac{\pi}{8}.$

   So the whole expression becomes

   $\cos\frac{7\pi}{8} + \cos\frac{\pi}{8} - 2\sin^2\frac{\pi}{8}.$

   But $\cos\frac{7\pi}{8} + \cos\frac{\pi}{8} = 0$, so we get $-2\sin^2\frac{\pi}{8}$. That is still a constant, not a product of four factors.

   6. Re-examine the options for a match.

      Compute $-2\sin^2\frac{\pi}{8}$ numerically: $\sin\frac{\pi}{8} = \sin 22.5^\circ \approx 0.3827$, so $\sin^2 \approx 0.1464$, times $-2$ gives about $-0.2929$.

      Now check option (C): $4 \cos\frac{\pi}{16} \cos\frac{3\pi}{8} \cos\frac{9\pi}{16}$.

      $\cos\frac{\pi}{16} \approx \cos 11.25^\circ \approx 0.9808$,

      $\cos\frac{3\pi}{8} = \cos 67.5^\circ \approx 0.3827$,

      $\cos\frac{9\pi}{16} = \cos 101.25^\circ \approx -0.1951$.

      Product: $0.9808 \times 0.3827 \times (-0.1951) \approx -0.0732$, times 4 gives $-0.2928$. That matches!

      So option (C) is correct numerically. The trick is that the expression does simplify to a product, but only after using identities that involve the $-1$ in a non-obvious way.

   7. Derive the identity properly.

      Start from the original:

   $\cos\frac{7\pi}{8} + \cos\frac{\pi}{4} + \cos\frac{\pi}{8} - 1.$

   Write $\cos\frac{7\pi}{8} = -\cos\frac{\pi}{8}$ as before, so the sum of the two cosines is zero. That leaves

   $\cos\frac{\pi}{4} - 1.$

   Now use $1 - \cos\frac{\pi}{4} = 2\sin^2\frac{\pi}{8}$, so

   $\cos\frac{\pi}{4} - 1 = -2\sin^2\frac{\pi}{8}.$

   Next, express $\sin\frac{\pi}{8}$ in terms of cosines of double/half angles. Since $\sin\frac{\pi}{8} = \cos\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \cos\frac{3\pi}{8}$, we have

   $-2\sin^2\frac{\pi}{8} = -2\cos^2\frac{3\pi}{8}.$

   Now use the identity $\cos^2\theta = \frac{1+\cos2\theta}{2}$:

   $$-2\cos^2\frac{3\pi}{8} = -2 \cdot \frac{1+\cos\frac{3\pi}{4}}{2} = -1 - \cos\frac{3\pi}{4}. …$$

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7. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.A and C lie in $\left[0, \dfrac{\pi}{2}\right]$ and B lies in $[0, 2\pi]$. If $\tan A + 3\cos B + 6\sin C = 1; 3\tan A + \cos B + 4\sin C = 4; 5\tan A + 3\cos B - 8\sin C = -2$, then $B - 2A - C =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Solve the system for $\tan A$, $\cos B$, and $\sin C$ using elimination, then use the given ranges to find $A$, $B$, $C$ individually, and compute $B - 2A - C = \frac{\pi}{2}$.

   We have three equations in three unknowns: $\tan A$, $\cos B$, and $\sin C$. The ranges tell us which quadrant each angle lies in, so we can pick the correct sign when recovering the angle from its trigonometric value.

   Let

   $x = \tan A$,

   $y = \cos B$,

   $z = \sin C$.

   The system becomes:

   1. $x + 3y + 6z = 1$
   2. $3x + y + 4z = 4$
   3. $5x + 3y - 8z = -2$

   We solve this linear system.

   Step 1: Eliminate $y$.

   Subtract equation (1) from equation (3):

   $(5x + 3y - 8z) - (x + 3y + 6z) = -2 - 1$

   $4x - 14z = -3$

   So $4x - 14z = -3$ …(4)

   Step 2: Eliminate $y$ again using (1) and (2).

   Multiply (1) by 3: $3x + 9y + 18z = 3$

   Subtract (2) from this: $(3x + 9y + 18z) - (3x + y + 4z) = 3 - 4$

   $8y + 14z = -1$ …(5)

   Step 3: Solve for $x$ and $z$.

   From (4): $4x = 14z - 3 \implies x = \frac{14z - 3}{4}$.

   Now use (5) to find $y$ in terms of $z$: $8y = -1 - 14z \implies y = \frac{-1 - 14z}{8}$.

   Substitute $x$ and $y$ into (1):

   $\frac{14z - 3}{4} + 3\left(\frac{-1 - 14z}{8}\right) + 6z = 1$

   Multiply through by 8:

   $2(14z - 3) + 3(-1 - 14z) + 48z = 8$

   $28z - 6 - 3 - 42z + 48z = 8$

   $(28z - 42z + 48z) - 9 = 8$

   $34z - 9 = 8$

   $34z = 17$

   $z = \frac{1}{2}$

   Step 4: Back-substitute.

   $x = \frac{14(\frac12) - 3}{4} = \frac{7 - 3}{4} = 1$

   $y = \frac{-1 - 14(\frac12)}{8} = \frac{-1 - 7}{8} = -1$

   So $\tan A = 1$, $\cos B = -1$, $\sin C = \frac12$.

   Step 5: Recover the angles from the given ranges. …

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8. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $\dfrac{2\sin\theta}{1+\cos\theta+\sin\theta}=y$, then $\dfrac{1-\cos\theta+\sin\theta}{1+\sin\theta}=$ (A) $y$ (B) $\dfrac{1}{y}$ (C) $1-y$ (D) $1+y$

   ›Reveal solutionSolution

   The key idea is to simplify the given expression by rationalizing or rewriting it in terms of $y$ using algebraic manipulation. The final result is that the expression equals $y$, so the correct option is (A).

   We start with the given relation:

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

   We need to find an expression for

   $\frac{1-\cos\theta+\sin\theta}{1+\sin\theta}.$

   The trick here is to notice that the numerator $1-\cos\theta+\sin\theta$ looks like it could be related to the denominator of $y$ after some algebraic juggling. A common approach is to multiply numerator and denominator of the target expression by something that brings in $2\sin\theta$, or to rewrite $1-\cos\theta$ using half-angle identities. But let’s keep it purely algebraic and avoid heavy trigonometry.

   1. Rewrite the target expression

      Let $T = \dfrac{1-\cos\theta+\sin\theta}{1+\sin\theta}$.

      We want to see if $T$ simplifies to $y$, $\frac{1}{y}$, $1-y$, or $1+y$.

   2. Multiply numerator and denominator of $T$ by something useful

      Notice that the denominator of $y$ is $1+\cos\theta+\sin\theta$. If we multiply $T$’s numerator and denominator by $(1+\cos\theta+\sin\theta)$, we might create a link.

      So consider:

   $T = \frac{1-\cos\theta+\sin\theta}{1+\sin\theta} \times \frac{1+\cos\theta+\sin\theta}{1+\cos\theta+\sin\theta}.$

   This gives:

   $T = \frac{(1-\cos\theta+\sin\theta)(1+\cos\theta+\sin\theta)}{(1+\sin\theta)(1+\cos\theta+\sin\theta)}.$

   3. Simplify the numerator The numerator is of the form $(A+B)(A-B)$? Not exactly. Let’s group terms: Let $A = 1+\sin\theta$ and $B = \cos\theta$. Then:

   $1-\cos\theta+\sin\theta = (1+\sin\theta) - \cos\theta = A - B,$

   and

   $1+\cos\theta+\sin\theta = (1+\sin\theta) + \cos\theta = A + B.$

   So the numerator becomes $(A-B)(A+B) = A^2 - B^2 = (1+\sin\theta)^2 - \cos^2\theta$.

   4. Expand and simplify

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

   Recall $\sin^2\theta - \cos^2\theta = -(\cos^2\theta - \sin^2\theta) = -\cos 2\theta$, but better: use $\sin^2\theta + \cos^2\theta = 1$.

   So:

   $$$$

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9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If $\cosh x = \dfrac{4}{3}$, then $3 \cosh x + 3^2 \cosh 2x + 3^3 \cosh 3x =$ (A) 175 (B) 81 (C) 64 (D) 27

   ›Reveal solutionSolution

   The key idea is to express $\cosh nx$ in terms of $\cosh x$ using the recurrence $\cosh(nx)=2\cosh x\cosh((n-1)x)-\cosh((n-2)x)$, then substitute $\cosh x=4/3$ and compute the sum step by step. The final result is 175.

   We are given $\cosh x = \frac{4}{3}$ and need to evaluate

   $3\cosh x + 3^2\cosh 2x + 3^3\cosh 3x$.

   ---

   Concept and Intuition

   The hyperbolic cosine satisfies the same double-angle and addition formulas as the ordinary cosine, but with no sign changes. In particular,

   $\cosh 2x = 2\cosh^2 x - 1$ and

   $\cosh 3x = 4\cosh^3 x - 3\cosh x$.

   Since we know $\cosh x$ exactly, we can compute $\cosh 2x$ and $\cosh 3x$ directly, then plug into the sum. This avoids any need to find $x$ itself.

   ---

   Step-by-step solution

   1. Compute $\cosh 2x$ Using the identity $\cosh 2x = 2\cosh^2 x - 1$:

   $\cosh 2x = 2\left(\frac{4}{3}\right)^2 - 1 = 2\cdot\frac{16}{9} - 1 = \frac{32}{9} - \frac{9}{9} = \frac{23}{9}.$

   2. Compute $\cosh 3x$ Using $\cosh 3x = 4\cosh^3 x - 3\cosh x$:

   $\cosh 3x = 4\left(\frac{4}{3}\right)^3 - 3\left(\frac{4}{3}\right) = 4\cdot\frac{64}{27} - 4 = \frac{256}{27} - \frac{108}{27} = \frac{148}{27}.$

   (Check: $4\cdot 64/27 = 256/27$, and $3\cdot 4/3 = 4 = 108/27$, so difference is $148/27$.)

   3. Evaluate each term of the sum
      • First term: $3\cosh x = 3 \cdot \frac{4}{3} = 4$. …

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10. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.$a_1, a_2, a_3$ are in arithmetic progression in which common difference is 2 and the first term is 2. If $\cos a_1 + \cos a_2 + \cos a_3 = \dfrac{\sin \alpha}{\sin \beta} \cos \gamma$ then $\alpha + \beta =$ (A) $2\gamma$ (B) $\gamma$ (C) $\dfrac{\gamma}{2}$ (D) $\gamma + 2$

    ›Reveal solutionSolution

    The sum of three cosines of an arithmetic progression simplifies using the sum-to-product identity, leading to a relation between the angles in the given expression; the answer is that $\alpha + \beta = \gamma$.

    We are told $a_1, a_2, a_3$ are in arithmetic progression with first term $2$ and common difference $2$. So:

    $a_1 = 2,\quad a_2 = 4,\quad a_3 = 6.$

    The problem gives:

    $\cos a_1 + \cos a_2 + \cos a_3 = \frac{\sin \alpha}{\sin \beta} \cos \gamma.$

    We need to find $\alpha + \beta$ in terms of $\gamma$ (or a constant).

    ---

    1. Compute the left-hand side directly

    $\cos 2 + \cos 4 + \cos 6.$

    These are in radians (since no degree symbol). But we can sum them using a known identity.

    2. Use the sum-to-product identity for three terms in AP

    For three angles in arithmetic progression: $x-d,\, x,\, x+d$, we have:

    $\cos(x-d) + \cos x + \cos(x+d) = \cos x + 2\cos x \cos d = \cos x\,(1 + 2\cos d).$

    Here $x = 4$, $d = 2$, so:

    $\cos 2 + \cos 4 + \cos 6 = \cos 4\,(1 + 2\cos 2).$

    3. Simplify $1 + 2\cos 2$ using a double-angle identity

    Recall $\cos 2\theta = 2\cos^2\theta - 1$, so $1 + 2\cos 2 = 1 + 2(2\cos^2 1 - 1) = 1 + 4\cos^2 1 - 2 = 4\cos^2 1 - 1$.

    But that doesn’t look like a simple ratio yet. Instead, use the identity:

    $1 + 2\cos 2 = \frac{\sin 3}{\sin 1}.$

    Check: $\sin 3 = \sin(2+1) = \sin 2\cos 1 + \cos 2\sin 1$, and $\sin 2 = 2\sin 1\cos 1$, so:

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

    But $2\cos^2 1 = 1 + \cos 2$, so $2\cos^2 1 + \cos 2 = 1 + 2\cos 2$. Hence $\sin 3 = \sin 1 (1 + 2\cos 2)$, so indeed:

    $1 + 2\cos 2 = \frac{\sin 3}{\sin 1}.$

    4. Substitute back

    $\cos 2 + \cos 4 + \cos 6 = \cos 4 \cdot \frac{\sin 3}{\sin 1}.$

    So the left-hand side equals:

    $\frac{\sin 3}{\sin 1} \cos 4.$ …

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11. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.If $\cos A + \cos B + \cos C = 0 = \sin A + \sin B + \sin C$, then $\cos(A-B) + \cos(B-C) + \cos(C-A) =$ (A) $0$ (B) $\dfrac{1}{2}$ (C) $\dfrac{3}{2}$ (D) $-\dfrac{3}{2}$

    ›Reveal solutionSolution

    The condition means the three vectors (cos A, sin A), (cos B, sin B), (cos C, sin C) sum to zero, so they form an equilateral triangle on the unit circle; the required sum equals $3/2$, so the answer is (C).

    We are given that

    $\cos A + \cos B + \cos C = 0 \quad\text{and}\quad \sin A + \sin B + \sin C = 0.$

    We need the value of

    $\cos(A-B) + \cos(B-C) + \cos(C-A).$

    ---

    Concept and Intuition

    Think of each pair $(\cos\theta, \sin\theta)$ as a point on the unit circle — that is, as a vector of length 1. The condition says that the sum of three such vectors is the zero vector. When three equal-length vectors sum to zero, they must be spaced equally around the circle: they are the vertices of an equilateral triangle centered at the origin. That means the angles differ by $120^\circ$ (or $2\pi/3$ radians). Then each cosine of a difference like $\cos(A-B)$ becomes $\cos(120^\circ) = -1/2$, and the sum of three such terms is $-3/2$. Wait — that would give $-3/2$, but we must check carefully: the problem asks for $\cos(A-B) + \cos(B-C) + \cos(C-A)$, and if the angles are equally spaced, each difference is $\pm 120^\circ$, giving $\cos 120^\circ = -1/2$, so the sum would be $-3/2$. That is option (D). But is that correct? Let’s test with a concrete example: take $A=0^\circ$, $B=120^\circ$, $C=240^\circ$. Then $\cos A + \cos B + \cos C = 1 - 1/2 - 1/2 = 0$, and $\sin A + \sin B + \sin C = 0 + \sqrt{3}/2 - \sqrt{3}/2 = 0$. Then $\cos(A-B) = \cos(-120^\circ) = -1/2$, $\cos(B-C) = \cos(-120^\circ) = -1/2$, $\cos(C-A) = \cos(240^\circ) = -1/2$. Sum = $-3/2$. So indeed the sum is $-3/2$. But wait — the options include both $3/2$ and $-3/2$. So the answer is (D). However, let’s verify algebraically to be sure there is no sign trick.

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    Step-by-step algebraic derivation

    1. Square and add the given equations.

    $(\cos A + \cos B + \cos C)^2 + (\sin A + \sin B + \sin C)^2 = 0^2 + 0^2 = 0.$

    Expand each square:

    $\cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A\cos B + \cos B\cos C + \cos C\cos A) = 0,$

    $\sin^2 A + \sin^2 B + \sin^2 C + 2(\sin A\sin B + \sin B\sin C + \sin C\sin A) = 0.$

    Add them:

    $$(\cos^2 A+\sin^2 A) + (\cos^2 B+\sin^2 B) + (\cos^2 C+\sin^2 C) + 2\big[(\cos A\cos B+\sin A\sin B) + \dots\big] = 0. …$$

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12. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.If $\theta$ is an acute angle, $x = \sum_{n=0}^{\infty} \cos^{2n} \theta$, $y = \sum_{n=0}^{\infty} \sin^{2n} \theta$ then $\frac{1}{x^2} + \frac{1}{y^2} =$ (A) $1 - \frac{2}{xy}$ (B) $1 - \frac{1}{xy}$ (C) $1 + \frac{1}{xy}$ (D) $1 + \frac{2}{xy}$

    ›Reveal solutionSolution

    Summing the geometric series gives $x=\csc^2\theta$ and $y=\sec^2\theta$, so $\dfrac1{x^2}+\dfrac1{y^2}=\sin^4\theta+\cos^4\theta = 1-\dfrac{2}{xy}$ — option (A).

    Both are geometric series with common ratio strictly less than $1$ for acute $\theta$:

    $x=\sum_{n=0}^\infty \cos^{2n}\theta=\frac{1}{1-\cos^2\theta}=\frac{1}{\sin^2\theta},\qquad y=\sum_{n=0}^\infty \sin^{2n}\theta=\frac{1}{1-\sin^2\theta}=\frac{1}{\cos^2\theta}.$

    Therefore

    $\frac{1}{x}=\sin^2\theta,\qquad \frac{1}{y}=\cos^2\theta.$

    Now

    $\frac{1}{x^2}+\frac{1}{y^2}=\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2\sin^2\theta\cos^2\theta.$ …

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