Q.If $\cos x = -\frac{3}{5}$, $x$ lies in the third quadrant, find the values of other five trigonometric functions.

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The number of real roots of the equation $x^7 + 3x^5 - 13x^3 - 15x = 0$ is (A) $5$ (B) $1$ (C) $7$ (D) $3$

   ›Reveal solutionSolution

   The equation factors as $x(x^6 + 3x^4 - 13x^2 - 15)=0$, and the sextic in $y=x^2$ has exactly one positive root, giving three real roots total. The correct option is (D).

   We start with the equation

   $x^7 + 3x^5 - 13x^3 - 15x = 0.$

   The key observation: every term contains a factor of $x$, so $x=0$ is an obvious root. The rest of the equation is even-powered in $x$, which suggests substituting $y = x^2$ to reduce the degree. This is a classic trick for polynomials with only odd or only even powers — it turns a high-degree equation into a lower-degree one in a new variable.

   1. Factor out $x$

   $x(x^6 + 3x^4 - 13x^2 - 15) = 0.$

   So $x=0$ is one real root. Now we need the number of real roots of

   $x^6 + 3x^4 - 13x^2 - 15 = 0.$

   2. Substitute $y = x^2$ Since $x^2 \ge 0$ for real $x$, let $y = x^2 \ge 0$. Then $x^4 = y^2$, $x^6 = y^3$, and the equation becomes

   $y^3 + 3y^2 - 13y - 15 = 0.$

   Every non-negative real root $y$ gives two real $x$-values ($x = \pm\sqrt{y}$), except $y=0$ which gives only $x=0$ (already counted).

   3. Analyze the cubic in $y$ Let $f(y) = y^3 + 3y^2 - 13y - 15$. We look for its real roots, especially non-negative ones.
      • $f(0) = -15 < 0$
      • $f(1) = 1 + 3 - 13 - 15 = -24 < 0$
      • $f(2) = 8 + 12 - 26 - 15 = -21 < 0$
      • $f(3) = 27 + 27 - 39 - 15 = 0$ So $y=3$ is a root. Factor it out:

   $y^3 + 3y^2 - 13y - 15 = (y-3)(y^2 + 6y + 5).$

   The quadratic $y^2 + 6y + 5 = (y+1)(y+5)$ has roots $y=-1$ and $y=-5$, both negative.

   4. Count the real $x$-roots The cubic has exactly one non-negative real root: $y=3$.
      • $y=3$ gives $x = \pm\sqrt{3}$, two distinct real roots.
      • Together with $x=0$, we have $1 + 2 = 3$ real roots.

   Watch out

   A common mistake is to forget that $y = x^2$ doubles the count for each positive $y$, or to miss the root $x=0$ entirely. Also, note that the cubic has three real roots (since its discriminant is positive), but two are negative and irrelevant for $x$ real.

   Tip

   Always check $x=0$ first when every term has a factor of $x$. Then use substitution $y=x^2$ to halve the degree — it turns a degree-7 equation into a cubic, which is much easier to analyze.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.In a triangle ABC, if $a = 2$, $\sin A = \dfrac{2}{3}$, $B = \dfrac{\pi}{3}$, then $\sqrt{5}b - 3c =$ (A) $-3$ (B) $3\sqrt{3}$ (C) $\sqrt{5} - \sqrt{3}$ (D) $2$

   ›Reveal solutionSolution

   Using the Law of Sines to find side lengths $b$ and $c$ from given $a$, $\sin A$, and angle $B$, then substituting into $\sqrt{5}b - 3c$ yields a simple numeric result.

   We are given a triangle $ABC$ with side $a = 2$ (opposite $A$), $\sin A = \frac{2}{3}$, and $B = \frac{\pi}{3}$. We need $\sqrt{5}b - 3c$.

   Concept & Intuition

   The Law of Sines relates sides to sines of opposite angles: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ (circumdiameter). Since we know $a$, $\sin A$, and $B$, we can find $b$ directly. To find $c$, we need $\sin C$, which comes from $C = \pi - A - B$. We know $\sin A$, so we can find $\cos A$ (using the Pythagorean identity, careful about sign — in a triangle, angles are between $0$ and $\pi$, and $\sin A$ positive means $A$ could be acute or obtuse, but here we check). Then compute $\sin C = \sin(\pi - A - B) = \sin(A+B)$. Finally, use the Law of Sines again for $c$.

   ---

   Step-by-step solution

   1. Find the common ratio $k$ from the Law of Sines

   $\frac{a}{\sin A} = \frac{2}{2/3} = 3.$

   So $k = 3$ is the constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 3$.

   2. Find $b$

   $b = 3 \sin B = 3 \cdot \sin\frac{\pi}{3} = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}.$

   3. Find $\cos A$ Since $\sin A = \frac{2}{3}$, we have

   $\cos^2 A = 1 - \sin^2 A = 1 - \frac{4}{9} = \frac{5}{9}.$

   So $\cos A = \pm \frac{\sqrt{5}}{3}$.

   In a triangle, if $A$ were obtuse ($>90^\circ$), $\cos A$ would be negative. But note: $B = 60^\circ$, so $A + B < 180^\circ$ always. Could $A > 90^\circ$? If $A > 90^\circ$, then $\sin A = 2/3$ is possible (e.g., $A \approx 138.2^\circ$). Then $A+B \approx 198.2^\circ > 180^\circ$, impossible. So $A$ must be acute. Hence $\cos A = +\frac{\sqrt{5}}{3}$.

   4. Find $\sin C$ $C = \pi - A - B$, so $\sin C = \sin(A+B)$.

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

   Substitute:

   $\sin C = \frac{2}{3} \cdot \cos\frac{\pi}{3} + \frac{\sqrt{5}}{3} \cdot \sin\frac{\pi}{3} = \frac{2}{3} \cdot \frac{1}{2} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{3}}{2} = \frac{1}{3} + \frac{\sqrt{15}}{6}.$

   Combine: $\frac{1}{3} = \frac{2}{6}$, so

   $\sin C = \frac{2 + \sqrt{15}}{6}.$

   5. Find $c$

   $c = 3 \sin C = 3 \cdot \frac{2 + \sqrt{15}}{6} = \frac{2 + \sqrt{15}}{2}.$

   6. Compute $\sqrt{5}b - 3c$

   $\sqrt{5}b = \sqrt{5} \cdot \frac{3\sqrt{3}}{2} = \frac{3\sqrt{15}}{2}.$

   $3c = 3 \cdot \frac{2 + \sqrt{15}}{2} = \frac{6 + 3\sqrt{15}}{2}.$

   Subtract:

   $\sqrt{5}b - 3c = \frac{3\sqrt{15}}{2} - \frac{6 + 3\sqrt{15}}{2} = \frac{3\sqrt{15} - 6 - 3\sqrt{15}}{2} = \frac{-6}{2} = -3.$

   Watch out

   A common mistake is forgetting to check whether $\cos A$ is positive or negative. If you blindly take $\cos A = +\frac{\sqrt{5}}{3}$, you get the correct answer here, but in other problems the sign matters — always verify that $A+B < 180^\circ$ to rule out an obtuse $A$.

   Tip

   Notice that the $\sqrt{15}$ terms cancel exactly — this is a hint that the problem was designed to yield a clean integer result.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $x \in \left(0, \dfrac{1}{\sqrt{2}}\right)$, then $\cot \left[ \cos^{-1} \left\{ \tan \left( \sin^{-1} x \right) \right\} \right] =$ (A) $\dfrac{\sqrt{1-x^2}}{\sqrt{1-2x^2}}$ (B) $\dfrac{x}{\sqrt{1-2x^2}}$ (C) $\dfrac{\sqrt{1-2x^2}}{\sqrt{1-x^2}}$ (D) $\dfrac{\sqrt{1-2x^2}}{x}$

   ›Reveal solutionSolution

   Unpack the nested inverse functions layer by layer to get $\dfrac{x}{\sqrt{1-2x^2}}$ — option (B).

   Evaluate $\cot\big[\cos^{-1}\{\tan(\sin^{-1}x)\}\big]$ for $x\in\left(0,\tfrac{1}{\sqrt2}\right)$; on this domain every intermediate angle is acute, so all roots are positive.

   Innermost. Let $\theta=\sin^{-1}x$, so $\sin\theta=x,\ \cos\theta=\sqrt{1-x^2}$ and

   $\tan(\sin^{-1}x)=\tan\theta=\frac{x}{\sqrt{1-x^2}}.$

   Middle. Let $\phi=\cos^{-1}\!\left(\dfrac{x}{\sqrt{1-x^2}}\right)$, so $\cos\phi=\dfrac{x}{\sqrt{1-x^2}}$ and

   $\sin\phi=\sqrt{1-\frac{x^2}{1-x^2}}=\sqrt{\frac{1-2x^2}{1-x^2}}.$

   Outermost.

   $\cot\phi=\frac{\cos\phi}{\sin\phi}=\frac{\dfrac{x}{\sqrt{1-x^2}}}{\sqrt{\dfrac{1-2x^2}{1-x^2}}}=\frac{x}{\sqrt{1-x^2}}\cdot\frac{\sqrt{1-x^2}}{\sqrt{1-2x^2}}=\frac{x}{\sqrt{1-2x^2}}.$

   Check at $x=\tfrac12$: this gives $\tfrac{0.5}{\sqrt{0.5}}\approx0.707$, and a direct numerical evaluation of $\cot[\cos^{-1}(\tan(\pi/6))]$ also gives $0.707$.

   ✓Final answer

   $\dfrac{x}{\sqrt{1-2x^2}}$ — option (B).

4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.One of the $5^{\text{th}}$ roots of $\omega$ is (A) $\text{cis}\left(\dfrac{4\pi}{3}\right)$ (B) $\text{cis}\left(\dfrac{7\pi}{15}\right)$ (C) $\text{cis}\left(\dfrac{11\pi}{15}\right)$ (D) $\text{cis}\left(\dfrac{9\pi}{15}\right)$

   ›Reveal solutionSolution

   With $\omega = \operatorname{cis}\tfrac{2\pi}{3}$ (a primitive cube root of unity), its fifth roots are $\operatorname{cis}\!\left(\tfrac{2\pi/3 + 2k\pi}{5}\right)$. For $k=3$ this gives $\operatorname{cis}\tfrac{4\pi}{3}$ — option (A).

   Step 1 — Identify $\omega$

   $\omega$ denotes the primitive cube root of unity:

   $\omega = e^{2\pi i/3} = \operatorname{cis}\frac{2\pi}{3}, \qquad |\omega|=1.$

   Step 2 — Fifth roots via De Moivre

   The fifth roots of $\operatorname{cis}\theta$ are

   $z_k = \operatorname{cis}\!\left(\frac{\theta + 2k\pi}{5}\right), \quad k=0,1,2,3,4,$

   with $\theta = \tfrac{2\pi}{3}$:

   $z_k = \operatorname{cis}\!\left(\frac{2\pi}{15} + \frac{2k\pi}{5}\right).$

   Step 3 — Evaluate for $k=3$

   $z_3 = \operatorname{cis}\!\left(\frac{2\pi}{15} + \frac{6\pi}{5}\right) = \operatorname{cis}\!\left(\frac{2\pi}{15} + \frac{18\pi}{15}\right) = \operatorname{cis}\frac{20\pi}{15} = \operatorname{cis}\frac{4\pi}{3}.$

   Step 4 — Verify

   $\left(\operatorname{cis}\frac{4\pi}{3}\right)^5 = \operatorname{cis}\frac{20\pi}{3} = \operatorname{cis}\!\left(\frac{20\pi}{3} - 6\pi\right) = \operatorname{cis}\frac{2\pi}{3} = \omega. \checkmark$

   The other options do not satisfy $z^5=\omega$ (e.g. $\left(\operatorname{cis}\tfrac{11\pi}{15}\right)^5 = \operatorname{cis}\tfrac{11\pi}{3} = \operatorname{cis}\tfrac{5\pi}{3} \ne \omega$).

   ✓Final answer

   A fifth root of $\omega$ is $\operatorname{cis}\dfrac{4\pi}{3}$ — option (A).

   ANSWER: A

5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The general solution of the equation $\sqrt{6-5\cos x+7\sin^2 x}-\cos x=0$ also satisfies the equation $\sqrt{6-5\cos x+7\sin^2 x}-\cos x=0$ (A) $\tan x + \cot x = 2$ (B) $\cot x + \csc x = 1$ (C) $\tan x + \sec x = 1$ (D) $\sec x + \csc x = 2$

   ›Reveal solutionSolution

   Solving the radical equation forces $\cos x = 1$ (so $x = 2n\pi$), and the only option consistent with this is $\tan x + \sec x = 1$.

   Solution

   Write the equation as $\sqrt{6-5\cos x+7\sin^2 x}=\cos x$, which requires $\cos x \ge 0$. Squaring and using $\sin^2 x = 1-\cos^2 x$:

   $6-5\cos x+7(1-\cos^2 x)=\cos^2 x$

   $13-5\cos x-7\cos^2 x=\cos^2 x$

   $8\cos^2 x+5\cos x-13=0$

   Factoring, $\cos x = \dfrac{-5\pm\sqrt{25+416}}{16}=\dfrac{-5\pm 21}{16}$, giving $\cos x = 1$ or $\cos x=-\dfrac{13}{8}$ (rejected).

   So $\cos x = 1$, i.e. $x = 2n\pi$, with $\sin x = 0$.

   Testing the options at $x=2n\pi$: options (A), (B), (D) involve $\cot x$ or $\csc x$, which are undefined when $\sin x = 0$. For (C): $\tan x + \sec x = 0 + 1 = 1$. $\checkmark$

   ✓Final answer

   $(C)\ \tan x + \sec x = 1$

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $2 \tanh^{-1} x = \sinh^{-1} \left( \dfrac{4}{3} \right)$ then $\cosh^{-1} \left( \dfrac{1}{x} \right) =$ (A) $\log(\sqrt{2} + 1)$ (B) $\log(\sqrt{2} - 1)$ (C) $\log(2 + \sqrt{3})$ (D) $\log(2 - \sqrt{3})$

   ›Reveal solutionSolution

   Solving gives $x=\dfrac{1}{2}$, so $\cosh^{-1}(1/x)=\cosh^{-1}2=\log(2+\sqrt{3})$.

   Solution

   Compute the right side: $\sinh^{-1}\!\left(\dfrac{4}{3}\right)=\log\!\left(\dfrac{4}{3}+\sqrt{\dfrac{16}{9}+1}\right)=\log\!\left(\dfrac{4}{3}+\dfrac{5}{3}\right)=\log 3.$

   The left side: $2\tanh^{-1}x=\log\dfrac{1+x}{1-x}$. Setting equal to $\log 3$:

   $\frac{1+x}{1-x}=3 \implies 1+x=3-3x \implies x=\frac{1}{2}.$

   Then $\dfrac{1}{x}=2$ and

   $\cosh^{-1}2=\log\!\left(2+\sqrt{2^2-1}\right)=\log(2+\sqrt{3}).$

   ✓Final answer

   $(C)\ \log(2+\sqrt{3})$

7. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $\tan\left(\dfrac{\pi}{4} + \dfrac{\alpha}{2}\right) = \tan^3\left(\dfrac{\pi}{4} + \dfrac{\beta}{2}\right)$, then $\dfrac{3 + \sin^2\beta}{1 + 3\sin^2\beta} =$ (A) $\dfrac{\cos\beta}{\cos\alpha}$ (B) $\dfrac{\cos\alpha}{\cos\beta}$ (C) $\dfrac{\sin\alpha}{\sin\beta}$ (D) $\dfrac{\cos^3\alpha}{\sin^3\beta}$

   ›Reveal solutionSolution

   With $R=\tan\!\big(\tfrac{\pi}{4}+\tfrac{\beta}{2}\big)$ the given relation yields $\sin\alpha=\dfrac{R^{6}-1}{R^{6}+1}$, $\sin\beta=\dfrac{R^{2}-1}{R^{2}+1}$, and the target reduces to $\dfrac{\sin\alpha}{\sin\beta}$ — option (C).

   Let $R=\tan\!\big(\tfrac{\pi}{4}+\tfrac{\beta}{2}\big)$ and $t=\tan\tfrac{\beta}{2}$, so $R=\dfrac{1+t}{1-t}$ and $t=\dfrac{R-1}{R+1}$. Writing $u=\tan\tfrac{\alpha}{2}$, the condition

   $\tan\!\Big(\tfrac{\pi}{4}+\tfrac{\alpha}{2}\Big)=\tan^{3}\!\Big(\tfrac{\pi}{4}+\tfrac{\beta}{2}\Big)=R^{3}$

   gives $\dfrac{1+u}{1-u}=R^{3}$, i.e. $u=\dfrac{R^{3}-1}{R^{3}+1}$.

   Using $\sin\theta=\dfrac{2\tan(\theta/2)}{1+\tan^{2}(\theta/2)}$:

   $\sin\beta=\frac{2t}{1+t^{2}}=\frac{R^{2}-1}{R^{2}+1},\qquad \sin\alpha=\frac{2u}{1+u^{2}}=\frac{R^{6}-1}{R^{6}+1}.$

   For the target, with $\sin^{2}\beta=\dfrac{4t^{2}}{(1+t^{2})^{2}}$,

   $\frac{3+\sin^{2}\beta}{1+3\sin^{2}\beta}=\frac{3(1+t^{2})^{2}+4t^{2}}{(1+t^{2})^{2}+12t^{2}}=\frac{3+10t^{2}+3t^{4}}{1+14t^{2}+t^{4}}.$

   Substituting $t=\dfrac{R-1}{R+1}$ and simplifying:

   $\frac{3+\sin^{2}\beta}{1+3\sin^{2}\beta}=\frac{R^{4}+R^{2}+1}{R^{4}-R^{2}+1}.$

   Now, using $R^{6}-1=(R^{2}-1)(R^{4}+R^{2}+1)$ and $R^{6}+1=(R^{2}+1)(R^{4}-R^{2}+1)$,

   $\frac{\sin\alpha}{\sin\beta}=\frac{R^{6}-1}{R^{6}+1}\cdot\frac{R^{2}+1}{R^{2}-1}=\frac{R^{4}+R^{2}+1}{R^{4}-R^{2}+1}.$

   The two expressions are identical.

   ✓Final answer

   $\dfrac{3+\sin^{2}\beta}{1+3\sin^{2}\beta}=\dfrac{\sin\alpha}{\sin\beta}$ — option (C).

   ANSWER: C

8. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.One of the values of $\left(\sqrt{3} - i\right)^{\frac{2}{3}}$ is (A) $2^{\frac{2}{5}} (1 - \sqrt{3}\ i)$ (B) $2^{-\frac{3}{5}} (\sqrt{3} + i)$ (C) $2^{\frac{2}{5}} (\sqrt{3} - i)$ (D) $2^{-\frac{3}{5}} (1 + \sqrt{3}\ i)$

   ›Reveal solutionSolution

   Write $\sqrt3 - i$ in polar form and apply De Moivre's theorem to the fractional power. One of the resulting values is $2^{-3/5}(1 + \sqrt3\,i)$, option (D).

   Step 1 — Polar form of $\sqrt3 - i$.

   Modulus: $r = \sqrt{(\sqrt3)^2 + (-1)^2} = 2$.

   Argument: $\cos\theta = \tfrac{\sqrt3}{2},\ \sin\theta = -\tfrac12$, so $\theta = -\tfrac{\pi}{6}$.

   Hence $\sqrt3 - i = 2\,e^{-i\pi/6}$.

   Step 2 — Apply the fractional power.

   By De Moivre's theorem the general value is

   $\left(2\,e^{-i\pi/6}\right)^{2/5} = 2^{2/5}\,e^{\,i\frac{2}{5}\left(-\frac{\pi}{6} + 2k\pi\right)}, \quad k = 0,1,2,3,4.$

   Step 3 — Pick the branch that matches an option.

   For $k = 3$ the argument is

   $\frac{2}{5}\left(-\frac{\pi}{6} + 6\pi\right) = -\frac{\pi}{15} + \frac{12\pi}{5} = \frac{\pi}{3},$

   giving

   $2^{2/5}\,e^{i\pi/3} = 2^{2/5}\left(\tfrac12 + \tfrac{\sqrt3}{2}i\right) = 2^{-3/5}(1 + \sqrt3\,i).$

   This matches option (D) (note $\left|2^{-3/5}(1+\sqrt3\,i)\right| = 2^{-3/5}\cdot 2 = 2^{2/5}$).

   ✓Final answer

   The correct option is (D).

9. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The value of $(1 - \cos \theta)(1 + \cos \theta)(1 + \cot^2 \theta)$ when $\theta = \dfrac{\pi}{15}$ is (A) $1$ (B) $\dfrac{1}{2}$ (C) $\dfrac{-1}{\sqrt{3}}$ (D) $2$

   ›Reveal solutionSolution

   The expression simplifies using fundamental trigonometric identities to a constant value, independent of $\theta$. The final value is $\boxed{1}$.

   The core idea here is to simplify the given trigonometric expression using fundamental identities before attempting to substitute the value of $\theta$. Often, in such problems, the specific value of the angle is a distractor if the expression simplifies to a constant.

   Let's break down the expression and simplify it step-by-step.

   1. Simplify the first part of the expression:

      We have $(1 - \cos \theta)(1 + \cos \theta)$. This is in the form $(a-b)(a+b)$, which simplifies to $a^2 - b^2$.

      Applying this algebraic identity:

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

   2. Apply a Pythagorean identity to the result:

      We know the fundamental Pythagorean identity:

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

      Rearranging this identity, we get $\sin^2 \theta = 1 - \cos^2 \theta$.

      So, the first part of the expression simplifies to $\sin^2 \theta$.

   3. Simplify the second part of the expression:

      The second part is $(1 + \cot^2 \theta)$. We can use another Pythagorean identity related to cotangent and cosecant.

      $1 + \cot^2 \theta = \csc^2 \theta$

      This identity can be derived from $\sin^2 \theta + \cos^2 \theta = 1$ by dividing all terms by $\sin^2 \theta$:

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

      $1 + \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \left(\frac{1}{\sin \theta}\right)^2$

      $1 + \cot^2 \theta = \csc^2 \theta$.

      So, the second part of the expression simplifies to $\csc^2 \theta$.

   4. Combine the simplified parts:

      Now, substitute the simplified forms back into the original expression:

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

      We know that $\csc \theta = \frac{1}{\sin \theta}$. Therefore, $\csc^2 \theta = \frac{1}{\sin^2 \theta}$.

      Substituting this:

      $(\sin^2 \theta)(\csc^2 \theta) = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta}$.

      Watch out

      This simplification is valid only if $\sin \theta \neq 0$. If $\sin \theta = 0$, then $\csc \theta$ is undefined, and the original expression would also be undefined due to $\cot^2 \theta$ (since $\cot \theta = \frac{\cos \theta}{\sin \theta}$).

      For $\theta = \frac{\pi}{15}$, $\sin \theta \neq 0$, so the simplification is valid.

   5. Final simplification:

      $\sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1$.

      The expression simplifies to $1$, regardless of the value of $\theta$ (as long as $\sin \theta \neq 0$). The given value $\theta = \frac{\pi}{15}$ is therefore irrelevant for finding the value of the simplified expression.

   The value of the expression is $1$.

   ✓Final answer

   The value of the expression is $\boxed{1}$.

10. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.If $x \neq y$ and $\sin x + \sin y = 3 (\cos y - \cos x)$, then $\tan(x-y) =$ (A) $\dfrac{\sqrt{3}}{2}$ (B) $-1$ (C) $\dfrac{3}{4}$ (D) $1$

    ›Reveal solutionSolution

    Use sum-to-product identities to rewrite the given equation, then simplify to a relation between $\tan\frac{x+y}{2}$ and $\tan\frac{x-y}{2}$, leading to $\tan(x-y) = \frac{3}{4}$.

    The key here is to transform the sum of sines and difference of cosines into products. When you see $\sin x + \sin y$ and $\cos y - \cos x$, your first instinct should be the sum-to-product formulas — they turn sums into products, which often lets you cancel common factors.

    1. Apply sum-to-product identities:

    $\sin x + \sin y = 2 \sin\frac{x+y}{2} \cos\frac{x-y}{2}$

    $\cos y - \cos x = -2 \sin\frac{x+y}{2} \sin\frac{y-x}{2}$

    Notice that $\sin\frac{y-x}{2} = -\sin\frac{x-y}{2}$, so:

    $\cos y - \cos x = -2 \sin\frac{x+y}{2} \cdot \left(-\sin\frac{x-y}{2}\right) = 2 \sin\frac{x+y}{2} \sin\frac{x-y}{2}$

    2. Substitute into the given equation $\sin x + \sin y = 3(\cos y - \cos x)$:

    $2 \sin\frac{x+y}{2} \cos\frac{x-y}{2} = 3 \cdot 2 \sin\frac{x+y}{2} \sin\frac{x-y}{2}$

    3. Since $x \neq y$, we have $\sin\frac{x+y}{2} \neq 0$? Actually, careful: if $\sin\frac{x+y}{2} = 0$, then both sides become zero, but the equation would hold trivially. However, the problem likely expects a non-trivial case. We can divide both sides by $2 \sin\frac{x+y}{2}$ (assuming it is non-zero; if it were zero, the equation would give $0=0$ and $\tan(x-y)$ would be undefined or arbitrary — but the options are finite, so we take the non-zero case). This gives:

    $\cos\frac{x-y}{2} = 3 \sin\frac{x-y}{2}$

    4. Therefore:

    $\tan\frac{x-y}{2} = \frac{1}{3}$

    5. Now use the double-angle formula for tangent:

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

    Watch out

    A common mistake is to misapply the formula for $\cos y - \cos x$. Remember: $\cos A - \cos B = -2 \sin\frac{A+B}{2} \sin\frac{A-B}{2}$. Here $A=y$, $B=x$, so $\cos y - \cos x = -2 \sin\frac{x+y}{2} \sin\frac{y-x}{2}$. Then note $\sin\frac{y-x}{2} = -\sin\frac{x-y}{2}$, giving the positive form used above.

    ✓Final answer

    The value is $\boxed{\dfrac{3}{4}}$, which corresponds to option (C).

11. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.If $\dfrac{2 \sin \alpha}{1 + \cos \alpha + \sin \alpha} = x$, then $\dfrac{1 - \cos \alpha - \sin \alpha}{\cos \alpha} =$ (A) $\dfrac{1}{x}$ (B) $-x$ (C) $1 - x$ (D) $1 + x$

    ›Reveal solutionSolution

    The expression equals $-x$ — option (B).

    Use the half-angle substitution $s=\sin\tfrac{\alpha}{2},\ c=\cos\tfrac{\alpha}{2}$, so $\sin\alpha=2sc$, $1+\cos\alpha=2c^{2}$, and $\cos\alpha=c^{2}-s^{2}$.

    Given quantity:

    $x = \frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha} = \frac{4sc}{2c^{2}+2sc} = \frac{4sc}{2c(c+s)} = \frac{2s}{c+s}.$

    Required quantity:

    $\frac{1-\cos\alpha-\sin\alpha}{\cos\alpha} = \frac{2s^{2}-2sc}{c^{2}-s^{2}} = \frac{2s(s-c)}{(c-s)(c+s)} = \frac{-2s(c-s)}{(c-s)(c+s)} = \frac{-2s}{c+s} = -x.$

    ✓Final answer

    $\dfrac{1-\cos\alpha-\sin\alpha}{\cos\alpha} = -x$ — option (B).

12. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.If $f(x) = -(\sin^2 x + \cos^5 x)$, then $\lim_{x \to 0} \frac{1}{x} f'(x)$ (A) exist and is equal to 0 (B) exist and is equal to 7 (C) exist and is equal to 3 (D) does not exist

    ›Reveal solutionSolution

    We first find the derivative $f'(x)$ and then evaluate the limit $\lim_{x \to 0} \frac{f'(x)}{x}$. This limit is of the indeterminate form $\frac{0}{0}$, which we resolve using L'Hopital's Rule. The limit exists and is equal to $\boxed{3}$.

    Concept and Intuition

    The problem asks for the limit of a ratio involving the derivative of a function. Our first step is to find the derivative $f'(x)$. Once we have $f'(x)$, we will substitute it into the limit expression $\lim_{x \to 0} \frac{f'(x)}{x}$.

    When we evaluate this limit by direct substitution, we will likely encounter an indeterminate form, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. These forms indicate that the limit cannot be found by simple substitution and require further techniques. The most common techniques for such forms are:

    1. Algebraic manipulation: Factoring, rationalizing, or using standard limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
    2. L'Hopital's Rule: If $\lim_{x \to a} \frac{g(x)}{h(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
    3. Taylor Series Expansion: Expanding the functions involved into their Maclaurin series (Taylor series around $x=0$) can often simplify the expression and reveal the limit, especially for trigonometric and exponential functions.

    For this specific problem, after finding $f'(x)$, we will see that direct substitution leads to the $\frac{0}{0}$ form. L'Hopital's Rule is a straightforward approach here, as differentiating the numerator and denominator once should resolve the indeterminacy.

    Let's proceed with finding $f'(x)$ and then evaluating the limit.

    Step-by-step solution

    1. Find the derivative $f'(x)$. The given function is $f(x) = -(\sin^2 x + \cos^5 x)$. We differentiate $f(x)$ with respect to $x$ using the chain rule and power rule:

    $f'(x) = - \left( \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^5 x) \right)$

    For the first term, $\frac{d}{dx}(\sin^2 x)$:
    Using the chain rule, $\frac{d}{dx}(u^2) = 2u \frac{du}{dx}$. Here $u = \sin x$.
    So, $\frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x$.
    We know the identity $2 \sin x \cos x = \sin(2x)$.
    Thus, $\frac{d}{dx}(\sin^2 x) = \sin(2x)$.
    
    For the second term, $\frac{d}{dx}(\cos^5 x)$:
    Using the chain rule, $\frac{d}{dx}(u^5) = 5u^4 \frac{du}{dx}$. Here $u = \cos x$.
    So, $\frac{d}{dx}(\cos^5 x) = 5 \cos^4 x \cdot \frac{d}{dx}(\cos x) = 5 \cos^4 x (-\sin x) = -5 \sin x \cos^4 x$.
    
    Substitute these derivatives back into the expression for $f'(x)$:
    

    $f'(x) = - (\sin(2x) - 5 \sin x \cos^4 x)$

    $f'(x) = 5 \sin x \cos^4 x - \sin(2x)$

    2. Set up the limit expression. We need to evaluate $\lim_{x \to 0} \frac{1}{x} f'(x)$, which is equivalent to $\lim_{x \to 0} \frac{f'(x)}{x}$. Substitute the expression for $f'(x)$ we just found:

    $\lim_{x \to 0} \frac{5 \sin x \cos^4 x - \sin(2x)}{x}$

    3. Check for indeterminate form.

       Let's evaluate the numerator and denominator as $x \to 0$:

       • As $x \to 0$, $\sin x \to 0$.
       • As $x \to 0$, $\cos x \to 1$, so $\cos^4 x \to 1^4 = 1$.
       • As $x \to 0$, $\sin(2x) \to \sin(0) = 0$.
       • The numerator approaches $5(0)(1) - 0 = 0$.
       • The denominator approaches $0$. This is an indeterminate form of type $\frac{0}{0}$.

    4. Apply L'Hopital's Rule.

       Since we have the $\frac{0}{0}$ form, we can apply L'Hopital's Rule. We differentiate the numerator and the denominator separately with respect to $x$.

       Let $N(x) = 5 \sin x \cos^4 x - \sin(2x)$ and $D(x) = x$.

       Then $N'(x) = \frac{d}{dx}(5 \sin x \cos^4 x - \sin(2x))$ and $D'(x) = \frac{d}{dx}(x) = 1$.

       To find $N'(x)$:

       • For the term $5 \sin x \cos^4 x$, we use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = 5 \sin x$ and $v = \cos^4 x$. Then $u' = \frac{d}{dx}(5 \sin x) = 5 \cos x$. And $v' = \frac{d}{dx}(\cos^4 x) = 4 \cos^3 x \cdot \frac{d}{dx}(\cos x) = 4 \cos^3 x (-\sin x) = -4 \sin x \cos^3 x$. So, $\frac{d}{dx}(5 \sin x \cos^4 x) = (5 \cos x)(\cos^4 x) + (5 \sin x)(-4 \sin x \cos^3 x)$

    $= 5 \cos^5 x - 20 \sin^2 x \cos^3 x$

    *   For the term $\sin(2x)$, we use the chain rule:
    

    $\frac{d}{dx}(\sin(2x)) = \cos(2x) \cdot \frac{d}{dx}(2x) = 2 \cos(2x)$

    Combining these, the derivative of the numerator $N'(x)$ is:
    

    $N'(x) = 5 \cos^5 x - 20 \sin^2 x \cos^3 x - 2 \cos(2x)$

    Now, apply L'Hopital's Rule:
    

    $\lim_{x \to 0} \frac{N(x)}{D(x)} = \lim_{x \to 0} \frac{N'(x)}{D'(x)}$

    $= \lim_{x \to 0} \frac{5 \cos^5 x - 20 \sin^2 x \cos^3 x - 2 \cos(2x)}{1}$

    Finally, substitute $x=0$ into this expression:
    *   As $x \to 0$, $\cos x \to 1$, so $\cos^5 x \to 1^5 = 1$.
    *   As $x \to 0$, $\sin x \to 0$, so $\sin^2 x \to 0^2 = 0$.
    *   As $x \to 0$, $\cos(2x) \to \cos(0) = 1$.
    
    Substituting these values:
    

    $5(1) - 20(0)(1) - 2(1) = 5 - 0 - 2 = 3$

    > [!TIP]
    > **Alternative using Taylor Series Expansion:**
    > We can also use Maclaurin series expansions for $\sin x$ and $\cos x$ around $x=0$:
    > $\sin x = x - \frac{x^3}{3!} + O(x^5)$
    > $\cos x = 1 - \frac{x^2}{2!} + O(x^4)$
    > $\sin(2x) = 2x - \frac{(2x)^3}{3!} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$
    >
    > From $f'(x) = 5 \sin x \cos^4 x - \sin(2x)$:
    > We need terms up to $x^3$ in $f'(x)$ to get a constant term after dividing by $x$.
    > $\cos^4 x = (1 - \frac{x^2}{2} + O(x^4))^4 \approx 1 - 4(\frac{x^2}{2}) + O(x^4) = 1 - 2x^2 + O(x^4)$ (using binomial expansion $(1+a)^n \approx 1+na$ for small $a$)
    >
    > Now substitute these into $f'(x)$:
    > $5 \sin x \cos^4 x \approx 5 \left(x - \frac{x^3}{6}\right) (1 - 2x^2)$
    > $\approx 5 \left(x - 2x^3 - \frac{x^3}{6} + O(x^5)\right)$
    > $\approx 5x - 10x^3 - \frac{5x^3}{6} + O(x^5)$
    >
    > So, $f'(x) \approx \left(5x - 10x^3 - \frac{5x^3}{6}\right) - \left(2x - \frac{4x^3}{3}\right)$
    > $f'(x) \approx (5x - 2x) + \left(-10 - \frac{5}{6} + \frac{4}{3}\right)x^3 + O(x^5)$
    > $f'(x) \approx 3x + \left(-\frac{60}{6} - \frac{5}{6} + \frac{8}{6}\right)x^3 + O(x^5)$
    > $f'(x) \approx 3x + \left(-\frac{57}{6}\right)x^3 + O(x^5) = 3x - \frac{19}{2}x^3 + O(x^5)$
    >
    > Then, $\lim_{x \to 0} \frac{f'(x)}{x} = \lim_{x \to 0} \frac{3x - \frac{19}{2}x^3 + O(x^5)}{x}$
    > $= \lim_{x \to 0} \left(3 - \frac{19}{2}x^2 + O(x^4)\right)$
    > $= 3$.
    > Both methods yield the same result.
    

    The limit exists and is equal to 3.

    ✓Final answer

    The limit $\lim_{x \to 0} \frac{1}{x} f'(x)$ exists and is equal to $\boxed{3}$.