Q.Prove that $\dfrac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \dfrac{1 + \sin A}{\cos A}$.

Showing the 12 most recent of 42 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\cos x = \tan y$, $\cos y = \tan z$ and $\cos z = \tan x$, then a value of $\sin x$ is (A) $2\cos 18°$ (B) $\cos 18°$ (C) $2\sin 18°$ (D) $\sin 18°$

   ›Reveal solutionSolution

   The cyclic system $\cos x=\tan y,\ \cos y=\tan z,\ \cos z=\tan x$ admits the symmetric solution $x=y=z$, which reduces to $\cos x=\tan x$; solving this quadratic in $\sin x$ gives $\sin x=2\sin18^\circ$.

   Concept and Intuition

   When a system of equations is cyclically symmetric in three variables (each equation is the same relation shifted by one variable), the natural first solution to look for is the symmetric one, $x=y=z$. Substituting this collapses three coupled equations into one, which is then a standard trigonometric equation solvable via the exact value of $\sin18^\circ$ from the regular pentagon.

   Step-by-Step Solution

   1. Assume the symmetric case $x=y=z$; each of the three given equations becomes $\cos x=\tan x$.
   2. Rewrite: $\cos x=\dfrac{\sin x}{\cos x} \Rightarrow \cos^2 x=\sin x \Rightarrow 1-\sin^2x=\sin x$.
   3. Rearranged: $\sin^2x+\sin x-1=0$.
   4. Solve the quadratic in $\sin x$: $\sin x=\dfrac{-1\pm\sqrt{1+4}}{2}=\dfrac{-1\pm\sqrt5}{2}$.
   5. Since $\sin x$ must lie in $[-1,1]$, take the positive root: $\sin x=\dfrac{\sqrt5-1}{2}$.
   6. Recall the classical exact value $\sin18^\circ=\dfrac{\sqrt5-1}{4}$, so $\sin x=2\left(\dfrac{\sqrt5-1}{4}\right)=2\sin18^\circ$.

   Common Mistakes

   • Taking the negative root of the quadratic, which falls outside the valid range for $\sin x$ derived from a positive quadratic expression equal to $\cos^2 x\ge0$ context (and doesn't match any option).
   • Confusing the exact values of $\sin18^\circ$ and $\cos36^\circ$ (they are related but numerically different: $\sin18^\circ\approx0.309$, $\cos36^\circ\approx0.809$).

   ✓Final answer

   The correct option is (C) — $2\sin 18°$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $a,b,c$ are non-zero real numbers and $\alpha, \beta$ are the values of $\theta$ satisfying the equation $a\cos\theta + b\sin\theta + c = 0$, then $\sec\alpha + \sec\beta =$ (A) $-\dfrac{2ac}{b^2-c^2}$ (B) $\dfrac{2ac}{b^2-c^2}$ (C) $\dfrac{2ab}{c^2-a^2}$ (D) $-\dfrac{2ab}{c^2-a^2}$

   ›Reveal solutionSolution

   Isolating and squaring the linear trig equation turns it into a quadratic in $\cos\theta$; applying Vieta's formulas to that quadratic gives $\sec\alpha+\sec\beta = \dfrac{2ac}{b^2-c^2}$.

   Concept and Intuition

   When an equation like $a\cos\theta+b\sin\theta+c=0$ has two roots $\alpha,\beta$ and we're asked for a symmetric function of $\cos\alpha,\cos\beta$ (like $\sec\alpha+\sec\beta$), the standard trick is to eliminate $\sin\theta$ by isolating it and squaring (using $\sin^2\theta=1-\cos^2\theta$), which converts the linear equation into a genuine quadratic in $\cos\theta$. Then $\cos\alpha,\cos\beta$ become exactly the two roots of that quadratic, and Vieta's formulas (sum and product of roots) hand us everything we need.

   Step-by-Step Solution

   1. From $a\cos\theta+b\sin\theta+c=0$, isolate: $b\sin\theta = -(a\cos\theta+c)$.
   2. Square both sides: $b^2\sin^2\theta = (a\cos\theta+c)^2 = a^2\cos^2\theta+2ac\cos\theta+c^2$.
   3. Replace $\sin^2\theta=1-\cos^2\theta$: $b^2(1-\cos^2\theta) = a^2\cos^2\theta+2ac\cos\theta+c^2$.
   4. Rearranging: $b^2 - b^2\cos^2\theta - a^2\cos^2\theta - 2ac\cos\theta - c^2 = 0$, i.e. $(a^2+b^2)\cos^2\theta + 2ac\cos\theta - (b^2-c^2) = 0$.
   5. So $\cos\alpha,\cos\beta$ are the roots of $(a^2+b^2)x^2+2ac\,x-(b^2-c^2)=0$. By Vieta's: $\cos\alpha+\cos\beta = \dfrac{-2ac}{a^2+b^2}$ and $\cos\alpha\cos\beta = \dfrac{-(b^2-c^2)}{a^2+b^2}=\dfrac{c^2-b^2}{a^2+b^2}$.
   6. $\sec\alpha+\sec\beta = \dfrac{1}{\cos\alpha}+\dfrac{1}{\cos\beta} = \dfrac{\cos\alpha+\cos\beta}{\cos\alpha\cos\beta} = \dfrac{-2ac/(a^2+b^2)}{(c^2-b^2)/(a^2+b^2)} = \dfrac{-2ac}{c^2-b^2} = \dfrac{2ac}{b^2-c^2}$.

   Common Mistakes

   • Forgetting that squaring can introduce extraneous roots, but here it's harmless since we only need $\cos\alpha,\cos\beta$ as roots of the resulting quadratic — the identity for their sum and product still holds for the genuine roots.
   • Sign error flipping $\dfrac{-2ac}{c^2-b^2}$ to $\dfrac{2ac}{b^2-c^2}$ — these are equal, but it's easy to drop a sign here.

   ✓Final answer

   The correct option is (B) — $\dfrac{2ac}{b^2-c^2}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $A + B + C = \pi$ and $\cos A = \cos B \cos C$, then $\tan A - \tan B - \tan C =$ (A) $-1$ (B) $0$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   The condition $\cos A=\cos B\cos C$ combined with $A+B+C=\pi$ forces $\tan B\tan C=2$, which in turn makes $\tan A=\tan B+\tan C$ exactly, giving $0$.

   Concept and Intuition

   Whenever $A+B+C=\pi$, angle $A$ is entirely determined by $B,C$ via $A=\pi-(B+C)$, so any trigonometric statement about $A$ can be converted into one about $B+C$ using supplementary-angle identities. The extra given condition on $\cos A$ then becomes an algebraic constraint purely between $\tan B$ and $\tan C$.

   Step-by-Step Solution

   1. From $A+B+C=\pi$: $A=\pi-(B+C)$, so $\cos A = \cos(\pi-(B+C)) = -\cos(B+C) = -(\cos B\cos C - \sin B\sin C) = -\cos B\cos C+\sin B\sin C$.
   2. Given $\cos A=\cos B\cos C$: equate to get $\cos B\cos C = -\cos B\cos C+\sin B\sin C \Rightarrow 2\cos B\cos C = \sin B\sin C$.
   3. Divide both sides by $\cos B\cos C$ (assumed nonzero): $2 = \tan B\tan C$.
   4. Now compute $\tan A = \tan(\pi-(B+C)) = -\tan(B+C) = -\dfrac{\tan B+\tan C}{1-\tan B\tan C} = -\dfrac{\tan B+\tan C}{1-2} = \tan B+\tan C$.
   5. Therefore $\tan A - \tan B - \tan C = 0$.

   Common Mistakes

   • Sign error expanding $\cos(\pi-(B+C))$ — it's $-\cos(B+C)$, not $+\cos(B+C)$.
   • Forgetting that $A+B+C=\pi$ alone (without the extra condition) gives the different identity $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ — here the extra condition changes the relationship entirely.

   ✓Final answer

   The correct option is (B) — $0$.

   ANSWER: B

4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.In a $\triangle ABC$, if $\cot\dfrac{A}{2} : \cot\dfrac{B}{2} : \cot\dfrac{C}{2} = 3 : 5 : 7$, then $\cos A : \cos B : \cos C =$ (A) $2 : 9 : 12$ (B) $6 : 5 : 4$ (C) $1 : 2 : 5$ (D) $3 : 4 : 5$

   ›Reveal solutionSolution

   Convert the cotangent ratio into a side ratio via $\cot(A/2)=(s-a)/r$, then use the law of cosines.

   Concept and Intuition

   Half-angle cotangents in a triangle are proportional to $s-a$, $s-b$, $s-c$ (the common factor $r$, the inradius, cancels in a ratio). This turns an angle-ratio problem into a much simpler linear side-ratio problem, after which the law of cosines gives the angle cosines directly.

   Step-by-Step Solution

   1. $\cot\dfrac{A}{2}=\dfrac{s-a}{r}$, $\cot\dfrac B2=\dfrac{s-b}{r}$, $\cot\dfrac C2=\dfrac{s-c}{r}$, so their ratio equals $(s-a):(s-b):(s-c)=3:5:7$.
   2. Let $s-a=3k,\ s-b=5k,\ s-c=7k$. Adding: $3s-(a+b+c)=15k$, but $a+b+c=2s$, so $3s-2s=15k \Rightarrow s=15k$.
   3. Then $a=s-3k=12k,\ b=s-5k=10k,\ c=s-7k=8k$, i.e. $a:b:c=12:10:8=6:5:4$.
   4. Take $a=12,b=10,c=8$ (any common scale). By the law of cosines: $\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{100+64-144}{160}=\dfrac{20}{160}=\dfrac18$.
   5. $\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{144+64-100}{192}=\dfrac{108}{192}=\dfrac{9}{16}$.
   6. $\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{144+100-64}{240}=\dfrac{180}{240}=\dfrac34$.
   7. $\cos A:\cos B:\cos C = \dfrac18:\dfrac9{16}:\dfrac34$. Multiply through by 16: $2:9:12$.

   Common Mistakes

   • Using $\cot(A/2)\propto a$ instead of $s-a$ (a common confusion with a different half-angle identity).
   • Sign/order slip when reading off which side is opposite which angle in the law of cosines.

   ✓Final answer

   The correct option is (A) — $2:9:12$.

   ANSWER: A

5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.In a $\triangle ABC$ if $a+c=5b$, then $\cot\dfrac{A}{2}\cot\dfrac{C}{2}=$ (A) $2$ (B) $\dfrac{1}{2}$ (C) $\dfrac{3}{2}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   A clean triangle identity turns $\cot\frac{A}{2}\cot\frac{C}{2}$ into $\frac{s}{s-b}$, and the given side relation makes this evaluate to $\frac{3}{2}$.

   Concept and Intuition

   In any triangle, $\tan\dfrac{A}{2}=\dfrac{r}{s-a}$ where $r$ is the inradius, so $\cot\dfrac{A}{2}=\dfrac{s-a}{r}$. Multiplying two such cotangents and using $r=\dfrac{\Delta}{s}$ (area over semi-perimeter) together with Heron's formula $\Delta^2=s(s-a)(s-b)(s-c)$ collapses things into a ratio purely of semi-perimeter terms.

   Step-by-Step Solution

   1. Use $\cot\dfrac{A}{2}=\dfrac{s-a}{r}$ and $\cot\dfrac{C}{2}=\dfrac{s-c}{r}$, so $\cot\dfrac{A}{2}\cot\dfrac{C}{2}=\dfrac{(s-a)(s-c)}{r^2}$.
   2. Since $r=\dfrac{\Delta}{s}$, we get $r^2=\dfrac{\Delta^2}{s^2}=\dfrac{s(s-a)(s-b)(s-c)}{s^2}=\dfrac{(s-a)(s-b)(s-c)}{s}$.
   3. So $\cot\dfrac{A}{2}\cot\dfrac{C}{2}=\dfrac{(s-a)(s-c)\cdot s}{(s-a)(s-b)(s-c)}=\dfrac{s}{s-b}$.
   4. Given $a+c=5b$, the semi-perimeter is $s=\dfrac{a+b+c}{2}=\dfrac{5b+b}{2}=3b$.
   5. Then $s-b=3b-b=2b$.
   6. So $\cot\dfrac{A}{2}\cot\dfrac{C}{2}=\dfrac{3b}{2b}=\dfrac{3}{2}$.

   Common Mistakes

   • Trying to expand $\cot\frac{A}{2}\cot\frac{C}{2}$ directly via the law of cosines rather than using the clean $\dfrac{s}{s-b}$ identity, which is far more error-prone.
   • Mixing up which side's "$s-b$" term appears — remember it's the side NOT among $A$ and $C$'s adjacent angle labels, i.e. $b$ is opposite angle $B$, distinct from $A,C$.

   ✓Final answer

   The correct option is (C) — $\dfrac{3}{2}$.

   ANSWER: C

6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If $A+B = \dfrac{\pi}{4}$, then $\dfrac{\cos B - \sin B}{\cos B + \sin B} =$ (A) $\sin A$ (B) $\cos A$ (C) $\tan A$ (D) $\cot A$

   ›Reveal solutionSolution

   This tests the tangent subtraction formula in disguise: dividing through by $\cos B$ turns the given ratio into $\tan(\pi/4-B)$, which is exactly $\tan A$.

   Concept and Intuition

   Whenever you see a ratio built from $\cos$ and $\sin$ of the same angle added/subtracted like $\cos B \pm \sin B$, dividing by $\cos B$ converts it into a tangent expression, because $\dfrac{1\pm\tan B}{1}$ is the numerator/denominator pattern of $\tan(45°\pm B)$. Recognising this pattern turns an awkward-looking trig ratio into a clean identity.

   Step-by-Step Solution

   1. Start with $\dfrac{\cos B-\sin B}{\cos B+\sin B}$.
   2. Divide numerator and denominator by $\cos B$ (valid since $\cos B\neq 0$ for the expression to be defined):

   $\frac{1-\tan B}{1+\tan B}$

   3. Recall $\tan\left(\dfrac{\pi}{4}-B\right)=\dfrac{\tan\frac{\pi}{4}-\tan B}{1+\tan\frac{\pi}{4}\tan B}=\dfrac{1-\tan B}{1+\tan B}$.
   4. So the given expression equals $\tan\left(\dfrac{\pi}{4}-B\right)$.
   5. Given $A+B=\dfrac{\pi}{4}$, so $A=\dfrac{\pi}{4}-B$.
   6. Hence the expression equals $\tan A$.

   Common Mistakes

   • Trying to expand using $A+B$ formulas directly on $\cos B, \sin B$ instead of recognizing the $\tan(45°\mp\theta)$ pattern.
   • Sign confusion between $\tan(\pi/4-B)$ and $\tan(\pi/4+B)$ — check by matching the numerator/denominator signs of the original ratio.

   ✓Final answer

   The correct option is (C) — $\tan A$.

   ANSWER: C

7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\tan\dfrac{2\pi}{7} \cdot \tan\dfrac{4\pi}{7} + \tan\dfrac{4\pi}{7} \cdot \tan\dfrac{\pi}{7} + \tan\dfrac{\pi}{7} \cdot \tan\dfrac{2\pi}{7} =$ (A) 7 (B) -7 (C) 3 (D) -3

   ›Reveal solutionSolution

   With $a=\pi/7$, the three angles $\pi/7,\,2\pi/7,\,4\pi/7$ satisfy $\tan(4\pi/7)=-\tan(3\pi/7)$; rewriting the target sum via $\tan(\pi/7),\tan(2\pi/7),\tan(3\pi/7)$ and their known symmetric-function values gives $-7$.

   Concept and Intuition

   $\tan\frac\pi7,\tan\frac{2\pi}7,\tan\frac{3\pi}7$ are the classic "sevenths" family: they are the positive roots of a cubic obtained from the multiple-angle expansion of $\tan 7\theta=0$, with known symmetric functions $\sum\tan^2=21,\ \sum\tan^2\tan^2=35,\ \prod\tan^2=7$ (so $\prod\tan=\sqrt7$). Since $\frac{4\pi}7=\pi-\frac{3\pi}7$, $\tan\frac{4\pi}7=-\tan\frac{3\pi}7$, letting us fold the given expression back into this familiar family.

   Step-by-Step Solution

   1. Let $a=\pi/7$, and set $t_1=\tan a,\ t_2=\tan2a,\ t_3=\tan3a$. Note $4a=\pi-3a$ so $\tan4a=-t_3$.
   2. Required sum $S=\tan2a\tan4a+\tan4a\tan a+\tan a\tan2a = t_2(-t_3)+(-t_3)t_1+t_1t_2 = t_1t_2-t_3(t_1+t_2)$.
   3. Numerically, $t_1=\tan25.71^\circ\approx0.48157$, $t_2=\tan51.43^\circ\approx1.25396$, $t_3=\tan77.14^\circ\approx4.38129$ (these are the standard roots of $\tan(7\theta)=0$ restricted to $\theta=\pi/7,2\pi/7,3\pi/7$, satisfying $t_1t_2t_3=\sqrt7$).
   4. $t_1t_2\approx0.60391$; $t_3(t_1+t_2)\approx4.38129\times1.73554\approx7.6038$.
   5. $S\approx0.60391-7.6038\approx-7.000$, an exact integer, confirming $S=-7$.

   Common Mistakes

   • Confusing $4\pi/7$ with an angle unrelated to the standard $\pi/7,2\pi/7,3\pi/7$ triple; the key move is $\tan(4\pi/7)=-\tan(3\pi/7)$.
   • Trying to force the identity "$\sum\tan=\prod\tan$" (valid for angles summing to $\pi$) onto a sum of products, which is a different symmetric function.

   ✓Final answer

   The correct option is (B) — -7.

   ANSWER: B

8. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If A and B are the values such that $(A+B)$ and $(A-B)$ are not odd multiples of $\dfrac{\pi}{2}$ and $2\tan(A+B) = 3\tan(A-B)$, then $\sin A \cos A =$ (A) $\sin B \cos B$ (B) $5\sin B \cos B$ (C) $\sin 2B$ (D) $\cos 2B$

   ›Reveal solutionSolution

   Converting the tangent condition into sines and cosines and applying product-to-sum identities reduces everything to a relation between $\sin2A$ and $\sin2B$, which directly gives $\sin A\cos A=5\sin B\cos B$.

   Concept and Intuition

   Whenever a condition is given in terms of $\tan(A+B)$ and $\tan(A-B)$ but the required conclusion is about $A$ and $B$ separately (via $\sin2A,\sin2B$), the trick is to write tangent as sine-over-cosine, cross-multiply, and use the product-to-sum identities $\sin X\cos Y=\tfrac12[\sin(X+Y)+\sin(X-Y)]$ — this naturally produces $\sin2A$ and $\sin2B$ since $(A+B)+(A-B)=2A$ and $(A+B)-(A-B)=2B$.

   Step-by-Step Solution

   1. $2\tan(A+B)=3\tan(A-B)\Rightarrow 2\dfrac{\sin(A+B)}{\cos(A+B)}=3\dfrac{\sin(A-B)}{\cos(A-B)}$, i.e. $2\sin(A+B)\cos(A-B)=3\sin(A-B)\cos(A+B)$ (valid since $A+B,A-B$ aren't odd multiples of $\pi/2$).
   2. Product-to-sum: $\sin(A+B)\cos(A-B)=\tfrac12[\sin2A+\sin2B]$ and $\sin(A-B)\cos(A+B)=\tfrac12[\sin2A-\sin2B]$.
   3. Substitute: $2\cdot\tfrac12[\sin2A+\sin2B]=3\cdot\tfrac12[\sin2A-\sin2B]$, i.e. $\sin2A+\sin2B=\tfrac32(\sin2A-\sin2B)$.
   4. Multiply by 2: $2\sin2A+2\sin2B=3\sin2A-3\sin2B\Rightarrow 5\sin2B=\sin2A$.
   5. So $\sin2A=5\sin2B\Rightarrow 2\sin A\cos A=5(2\sin B\cos B)\Rightarrow \sin A\cos A=5\sin B\cos B$.

   Common Mistakes

   • Errors in the sign when expanding $\sin(A-B)\cos(A+B)$ — it's easy to drop the minus sign on $\sin2B$ in that expansion.
   • Trying to solve for $A$ and $B$ individually instead of recognizing the whole point is to relate $\sin2A$ and $\sin2B$ directly.

   ✓Final answer

   The correct option is (B) — $5\sin B\cos B$.

   ANSWER: B

9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\tan\left(\dfrac{\pi}{4}+\alpha\right)=\tan^3\left(\dfrac{\pi}{4}+\beta\right)$, then $\tan(\alpha+\beta)\cot(\alpha-\beta)=$ (A) $\sec^2 2\beta+\tan^2 2\beta$ (B) $\csc^2 2\beta+\cot^2 2\beta$ (C) $2(\sec^2 2\beta+\tan^2 2\beta)$ (D) $4(\sec^2 2\beta+\tan^2 2\beta)$

   ›Reveal solutionSolution

   A substitution $A=\pi/4+\alpha$, $B=\pi/4+\beta$ turns the compound-angle expression into $\tan^2B+\cot^2B$-type algebra, and the key identity $\tan(\pi/4+\beta)=\sec2\beta+\tan2\beta$ finishes it. Answer: $2(\sec^22\beta+\tan^22\beta)$.

   Concept and Intuition

   Whenever an equation is given as $\tan(\text{something}+\alpha)=\tan^n(\text{something}+\beta)$, substituting $A,B$ for those combined angles converts a messy $\alpha,\beta$ expression into a clean function of $A,B$ alone — and $\alpha\pm\beta$ becomes $A\pm B$ shifted by a constant.

   Step-by-Step Solution

   1. Let $A=\dfrac{\pi}{4}+\alpha$, $B=\dfrac{\pi}{4}+\beta$. The condition becomes $\tan A=\tan^3B$; let $t=\tan B$, so $\tan A=t^3$.
   2. Note $\alpha=A-\pi/4$, $\beta=B-\pi/4$, so $\alpha+\beta=A+B-\pi/2$ and $\alpha-\beta=A-B$.
   3. $\tan(\alpha+\beta)=\tan\!\left(A+B-\dfrac\pi2\right)=-\cot(A+B)$, and $\cot(\alpha-\beta)=\cot(A-B)$.
   4. So the target is $-\cot(A+B)\cot(A-B)=-\dfrac{\cos(A+B)\cos(A-B)}{\sin(A+B)\sin(A-B)}$.
   5. Using product identities, $\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B$ and $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$. Substituting $\tan A=t^3,\ \tan B=t$ (so $\sin^2A=\tfrac{t^6}{1+t^6}$, $\cos^2A=\tfrac1{1+t^6}$, etc.) and simplifying yields

   $-\cot(A+B)\cot(A-B)=\frac{t^4+1}{t^2}=t^2+\frac1{t^2}.$

   6. Now use the identity $\tan\!\left(\dfrac\pi4+\beta\right)=\dfrac{\cos\beta+\sin\beta}{\cos\beta-\sin\beta}=\dfrac{1+\sin2\beta}{\cos2\beta}=\sec2\beta+\tan2\beta$, so $t=\sec2\beta+\tan2\beta$ and $\dfrac1t=\sec2\beta-\tan2\beta$.
   7. Then $t^2+\dfrac1{t^2}=(\sec2\beta+\tan2\beta)^2+(\sec2\beta-\tan2\beta)^2=2(\sec^22\beta+\tan^22\beta)$.

   Common Mistakes

   • Sign error converting $\tan(x-\pi/2)$ — remember $\tan(x-\pi/2)=-\cot x$, not $+\cot x$.
   • Forgetting the neat identity $\tan(\pi/4+\beta)=\sec2\beta+\tan2\beta$ and trying to grind through $t^2+1/t^2$ in raw $\sin\beta,\cos\beta$ terms (much messier).

   ✓Final answer

   The correct option is (C) — $2(\sec^2 2\beta+\tan^2 2\beta)$.

   ANSWER: C

10. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.Assertion (A): If $A = 10°, B = 16°, C = 19°$, then $\tan 2A \tan 2B + \tan 2B \tan 2C + \tan 2C \tan 2A = 1$. Reason (R): If $A+B+C = 180°$, $\cot\dfrac{A}{2} + \cot\dfrac{B}{2} + \cot\dfrac{C}{2} = \cot\dfrac{A}{2}\cot\dfrac{B}{2}\cot\dfrac{C}{2}$. Which of the following is correct? (A) Both (A) and (R) are true and (R) is the correct explanation of (A) (B) Both (A) and (R) are true and (R) is NOT correct explanation of (A) (C) (A) is true, (R) is false (D) (A) is false, (R) is true

    ›Reveal solutionSolution

    Both statements are true, and they are the same identity for angles summing to $90^\circ$ (one written in $\tan$, one in $\cot$) — so (R) genuinely explains (A).

    Concept and Intuition

    Everything here flows from one fact: if $X+Y+Z=90^\circ$ then $\tan X\tan Y+\tan Y\tan Z+\tan Z\tan X=1$.

    Why: $X+Y=90^\circ-Z$, so

    $\tan(X+Y)=\tan(90^\circ-Z)=\cot Z=\frac{1}{\tan Z} \ \Longrightarrow\ \frac{\tan X+\tan Y}{1-\tan X\tan Y}=\frac{1}{\tan Z},$

    and cross-multiplying gives $\tan X\tan Z+\tan Y\tan Z=1-\tan X\tan Y$, i.e. the identity.

    Now divide that identity throughout by $\tan X\tan Y\tan Z$:

    $\cot Z+\cot X+\cot Y=\cot X\cot Y\cot Z .$

    So the "tan-pairs sum to 1" identity and the "cot-sum = cot-product" identity are two faces of the same result, and the second is exactly the Reason (whose half-angles $\tfrac A2+\tfrac B2+\tfrac C2=90^\circ$ when $A+B+C=180^\circ$).

    Step-by-Step Solution

    1. Test the Assertion. $A=10^\circ,B=16^\circ,C=19^\circ$, so

    $2A+2B+2C=20^\circ+32^\circ+38^\circ=90^\circ .$

    With $X=2A,\ Y=2B,\ Z=2C$ summing to $90^\circ$, the identity above gives

    $\tan 2A\tan 2B+\tan 2B\tan 2C+\tan 2C\tan 2A=1 . \qquad\textbf{(A) is TRUE}$

    2. Test the Reason. If $A+B+C=180^\circ$ then $\dfrac A2+\dfrac B2+\dfrac C2=90^\circ$; applying the cotangent form of the same identity to these three half-angles,

    $\cot\frac A2+\cot\frac B2+\cot\frac C2=\cot\frac A2\cot\frac B2\cot\frac C2 . \qquad\textbf{(R) is TRUE}$

    3. Does (R) explain (A)? Take the triangle whose angles are $4A=40^\circ,\ 4B=64^\circ,\ 4C=76^\circ$ (sum $=180^\circ$ ✓). Its half-angles are $2A,2B,2C$, so (R) gives

    $\cot 2A+\cot 2B+\cot 2C=\cot 2A\cot 2B\cot 2C,$

    and multiplying through by $\tan 2A\tan 2B\tan 2C$ returns precisely the Assertion. So (R) is the correct explanation.

    Common Mistakes

    • Concluding (R) is irrelevant because it is stated in $\cot$ and half-angles while (A) uses $\tan$ of doubled angles — they are algebraically the same identity.
    • Trying to verify the assertion numerically with a calculator and rounding badly, instead of noticing the exact sum $2A+2B+2C=90^\circ$.
    • Confusing the $90^\circ$ identity with the $180^\circ$ one ($\tan X+\tan Y+\tan Z=\tan X\tan Y\tan Z$).

    ✓Final answer

    The correct option is (A) — Both (A) and (R) are true and (R) is the correct explanation of (A).

    ANSWER: A

11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $\sin x + \sin y = \alpha$, $\cos x + \cos y = \beta$ then $\mathrm{cosec}(x+y) =$ (A) $\dfrac{\beta^2 - \alpha^2}{\beta^2+\alpha^2}$ (B) $\dfrac{2\alpha\beta}{\beta^2 - \alpha^2}$ (C) $\dfrac{\beta^2+\alpha^2}{2\alpha\beta}$ (D) $\dfrac{2\alpha\beta}{\beta^2+\alpha^2}$

    ›Reveal solutionSolution

    Sum-to-product identities turn $\alpha,\beta$ into products involving $\frac{x+y}{2}$ and $\frac{x-y}{2}$; combining them eliminates the $\frac{x-y}{2}$ factor and isolates $\sin(x+y)$, giving $\text{cosec}(x+y)=\dfrac{\alpha^2+\beta^2}{2\alpha\beta}$.

    Concept and Intuition

    When a problem gives $\sin x+\sin y$ and $\cos x+\cos y$, the sum-to-product formulas convert both into a common factor of $\cos\frac{x-y}{2}$ times a sine or cosine of $\frac{x+y}{2}$. Squaring-and-adding eliminates $\frac{x+y}{2}$ (via $\sin^2+\cos^2=1$), while multiplying exploits the double-angle formula $2\sin A\cos A=\sin2A$ to bring in $\sin(x+y)$ directly. Combining both results cancels the leftover $\cos^2\frac{x-y}{2}$ factor.

    Step-by-Step Solution

    1. Sum-to-product: $\sin x+\sin y = 2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=\alpha$, and $\cos x+\cos y = 2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=\beta$.
    2. Let $A=\dfrac{x+y}{2}$, $B=\dfrac{x-y}{2}$, so $\alpha=2\sin A\cos B$, $\beta=2\cos A\cos B$.
    3. $\alpha^2+\beta^2 = 4\cos^2B(\sin^2A+\cos^2A) = 4\cos^2B$.
    4. $2\alpha\beta = 2\cdot(2\sin A\cos B)(2\cos A\cos B) = 8\sin A\cos A\cos^2B = 4\sin(2A)\cos^2B = 4\sin(x+y)\cos^2B$ (using $2A=x+y$).
    5. Divide: $\dfrac{2\alpha\beta}{\alpha^2+\beta^2} = \dfrac{4\sin(x+y)\cos^2B}{4\cos^2B} = \sin(x+y)$.
    6. So $\sin(x+y) = \dfrac{2\alpha\beta}{\alpha^2+\beta^2}$, and therefore $\text{cosec}(x+y) = \dfrac{\alpha^2+\beta^2}{2\alpha\beta}$.

    Common Mistakes

    • Forgetting the reciprocal step at the end (finding $\sin(x+y)$ but forgetting cosec is its reciprocal, giving an upside-down answer).
    • Sign or factor-of-2 slips in the double-angle expansion of $2\alpha\beta$.

    ✓Final answer

    The correct option is (C) — $\dfrac{\beta^2+\alpha^2}{2\alpha\beta}$.

    ANSWER: C

12. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.If A, B, C are the angles of a triangle, then $\dfrac{\sin A + \sin B + \sin C}{\sin^2\frac{A}{2} - \sin^2\frac{B}{2} + \sin^2\frac{C}{2} - 1} =$ (A) $-2\tan\dfrac{B}{2}$ (B) $-2\cot\dfrac{B}{2}$ (C) $2\tan\dfrac{B}{2}$ (D) $2\cot\dfrac{B}{2}$

    ›Reveal solutionSolution

    A triangle-identity simplification using $\sin A+\sin B+\sin C=4\cos\frac A2\cos\frac B2\cos\frac C2$ and half-angle manipulation of the denominator; the ratio equals $-2\cot\dfrac B2$.

    Concept and Intuition

    In any triangle, $A+B+C=\pi$, so half-angle sums like $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2$ are the key relation that turns sums/differences of half-angle sines into cosines and vice versa. The numerator has the classical identity $\sin A+\sin B+\sin C=4\cos\frac A2\cos\frac B2\cos\frac C2$; the denominator can be massaged into a similar product using $\sin^2X-\sin^2Y=\sin(X+Y)\sin(X-Y)$.

    Step-by-Step Solution

    1. Numerator: $\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$ (standard triangle identity).
    2. Denominator: write $\sin^2\frac A2-\sin^2\frac B2=\sin\left(\dfrac{A+B}2\right)\sin\left(\dfrac{A-B}2\right)$. Since $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2$, $\sin\left(\dfrac{A+B}2\right)=\cos\dfrac C2$. So $\sin^2\frac A2-\sin^2\frac B2=\cos\dfrac C2\sin\dfrac{A-B}2$.
    3. Also $\sin^2\dfrac C2-1=-\cos^2\dfrac C2$.
    4. So denominator $=\cos\dfrac C2\sin\dfrac{A-B}2-\cos^2\dfrac C2=\cos\dfrac C2\left[\sin\dfrac{A-B}2-\cos\dfrac C2\right]$.
    5. Since $\cos\dfrac C2=\sin\dfrac{A+B}2$, the bracket is $\sin\dfrac{A-B}2-\sin\dfrac{A+B}2=2\cos\dfrac A2\sin\left(-\dfrac B2\right)=-2\cos\dfrac A2\sin\dfrac B2$ (using $\sin X-\sin Y=2\cos\frac{X+Y}2\sin\frac{X-Y}2$).
    6. So denominator $=-2\cos\dfrac A2\cos\dfrac C2\sin\dfrac B2$.
    7. Ratio $=\dfrac{4\cos\frac A2\cos\frac B2\cos\frac C2}{-2\cos\frac A2\cos\frac C2\sin\frac B2}=-2\cdot\dfrac{\cos\frac B2}{\sin\frac B2}=-2\cot\dfrac B2$.

    Common Mistakes

    • Forgetting the sign when converting $\sin^2\frac C2-1$ to $-\cos^2\frac C2$.
    • Mixing up which half-angle survives in the final cotangent — a numeric check (e.g. a 3-4-5 triangle) confirms it is $B$'s half-angle.

    ✓Final answer

    The correct option is (B) — $-2\cot\dfrac B2$.

    ANSWER: B