Q.Express $\tan^{-1}\left(\dfrac{\cos x}{1-\sin x}\right)$, $-\dfrac{3\pi}{2} < x < \dfrac{\pi}{2}$ in the simplest form.

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $0<x<\pi$ and $\cos x+\sin x=\dfrac{1}{2}$ then $\tan x =$ (A) $\dfrac{4-\sqrt7}{3}$ (B) $\dfrac{4+\sqrt7}{3}$ (C) $\dfrac{-(4+\sqrt7)}{3}$ (D) $\dfrac{-4+\sqrt7}{3}$

   ›Reveal solutionSolution

   Squaring the given sum turns it into a symmetric-function problem; picking the sign consistent with $0<x<\pi$ gives $\tan x = -\dfrac{4+\sqrt7}{3}$.

   Concept and Intuition

   Given $\sin x+\cos x$, squaring produces $\sin x\cos x$ via $(\sin x+\cos x)^2=1+2\sin x\cos x$. Once both the sum and product of $\sin x,\cos x$ are known, they are the two roots of a quadratic — solve it and use the domain restriction to pick which root is $\sin x$ and which is $\cos x$.

   Step-by-Step Solution

   1. $(\cos x+\sin x)^2 = \frac14 \Rightarrow 1+2\sin x\cos x=\frac14 \Rightarrow \sin x\cos x = -\frac38$.
   2. $\sin x,\cos x$ are roots of $t^2-\frac12 t-\frac38=0$, i.e. $8t^2-4t-3=0$.
   3. $t = \dfrac{4\pm\sqrt{16+96}}{16}=\dfrac{4\pm\sqrt{112}}{16}=\dfrac{4\pm4\sqrt7}{16}=\dfrac{1\pm\sqrt7}{4}$.
   4. Since $0<x<\pi$, $\sin x>0$. Since $\sin x\cos x=-\frac38<0$, $\cos x$ must be negative. So $\sin x=\dfrac{1+\sqrt7}{4}$ (positive) and $\cos x=\dfrac{1-\sqrt7}{4}$ (negative, as $\sqrt7>1$). …

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2. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\dfrac{\cos x}{1+\sin x}+\tan x =$ (A) 1 (B) $\cos x + \sin x$ (C) $\sin^2 x$ (D) $\sec x$

   ›Reveal solutionSolution

   Rationalizing the fraction using $1-\sin^2x=\cos^2x$ turns it into $\sec x - \tan x$, which cancels the added $\tan x$ to leave $\sec x$.

   Concept and Intuition

   Expressions with $1+\sin x$ in the denominator are cleanly handled by multiplying by the conjugate $1-\sin x$, since $(1+\sin x)(1-\sin x)=1-\sin^2x=\cos^2x$ — this is the same trick used to rationalize surds, applied to trig.

   Step-by-Step Solution

   1. Multiply $\dfrac{\cos x}{1+\sin x}$ by $\dfrac{1-\sin x}{1-\sin x}$: numerator becomes $\cos x(1-\sin x)$, denominator becomes $1-\sin^2x=\cos^2x$.
   2. Simplify: $\dfrac{\cos x(1-\sin x)}{\cos^2 x} = \dfrac{1-\sin x}{\cos x} = \dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x} = \sec x - \tan x$. …

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3. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\tan x + \dfrac{\cos x}{1+\sin x} =$ (A) $\tan 2x$ (B) $\operatorname{cosec} x$ (C) $\sec x$ (D) $\cos 2x$

   ›Reveal solutionSolution

   Combine the two terms over a common denominator and use $\sin^2x+\cos^2x=1$ to collapse everything to $\sec x$.

   Concept and Intuition

   Whenever you see a sum of a trig ratio and a fraction like $\dfrac{\cos x}{1+\sin x}$, the standard move is to put both terms over the common denominator $\cos x(1+\sin x)$ and simplify the numerator using the Pythagorean identity.

   Step-by-Step Solution

   1. Write $\tan x = \dfrac{\sin x}{\cos x}$.
   2. Common denominator: $\dfrac{\sin x(1+\sin x) + \cos x\cdot\cos x}{\cos x(1+\sin x)} = \dfrac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1+\sin x)}$.
   3. Since $\sin^2x+\cos^2x=1$: numerator $= \sin x + 1$. …

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4. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.The value of $\dfrac{\sin\theta + \sin3\theta}{\cos\theta + \cos3\theta}$ is (A) $\cos2\theta$ (B) $\cot2\theta$ (C) $\tan2\theta$ (D) $\csc\theta + \sin\theta$

   ›Reveal solutionSolution

   Applying the sum-to-product identities to both numerator and denominator, the common factor $\cos\theta$ cancels, leaving $\tan2\theta$.

   Concept and Intuition

   Sum-to-product formulas convert a sum of sines (or cosines) at symmetric angles ($\theta$ and $3\theta$, straddling the midpoint $2\theta$) into a product involving that midpoint angle — a very common simplification trick.

   Step-by-Step Solution

   1. $\sin\theta + \sin3\theta = 2\sin\left(\dfrac{\theta+3\theta}{2}\right)\cos\left(\dfrac{3\theta-\theta}{2}\right) = 2\sin2\theta\cos\theta$.
   2. $\cos\theta + \cos3\theta = 2\cos\left(\dfrac{\theta+3\theta}{2}\right)\cos\left(\dfrac{3\theta-\theta}{2}\right) = 2\cos2\theta\cos\theta$. …

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5. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The general solution of $2\cos^2 x - 2\tan x + 1 = 0$ is (A) $n\pi + \dfrac{\pi}{4}, n \in Z$ (B) $2n\pi \pm \dfrac{\pi}{4}, n \in Z$ (C) $2n\pi \pm \dfrac{\pi}{3}, n \in Z$ (D) $n\pi \pm \dfrac{\pi}{3}, n \in Z$

   ›Reveal solutionSolution

   Substituting $t=\tan x$ converts the mixed $\cos^2x/\tan x$ equation into a cubic in $t$ with a unique real root $t=1$, giving $x=n\pi+\dfrac{\pi}{4}$.

   Concept and Intuition

   $\cos^2x$ and $\tan x$ can both be written purely in terms of $t=\tan x$ via $\cos^2x=\dfrac{1}{1+\tan^2x}$. This turns a trigonometric equation into an ordinary polynomial equation, which is easier to solve exactly.

   Step-by-Step Solution

   1. Substitute $\cos^2x=\dfrac{1}{1+t^2}$: $\dfrac{2}{1+t^2}-2t+1=0$.
   2. Multiply through by $(1+t^2)$: $2-2t(1+t^2)+(1+t^2)=0 \Rightarrow 2-2t-2t^3+1+t^2=0$.
   3. Rearranged: $-2t^3+t^2-2t+3=0$, i.e. $2t^3-t^2+2t-3=0$.
   4. Test small rational roots: $t=1$ gives $2(1)-1+2-3=0$ — a root.
   5. Divide out $(t-1)$: $2t^3-t^2+2t-3=(t-1)(2t^2+t+3)$.
   6. Discriminant of $2t^2+t+3$ is $1^2-4(2)(3)=1-24=-23<0$, so no further real roots. …

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6. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If $\sin\left(\dfrac{\pi}{4}\cot\theta\right) = \cos\left(\dfrac{\pi}{4}\tan\theta\right)$, then $\theta =$ (A) $2n\pi + \dfrac{\pi}{4}$ (B) $2n\pi \pm \dfrac{\pi}{4}$ (C) $2n\pi - \dfrac{\pi}{4}$ (D) $n\pi + \dfrac{\pi}{4}$

   ›Reveal solutionSolution

   Converting cosine to sine of the complementary angle and testing candidate solutions shows the general solution repeats every $\pi$, giving $\theta=n\pi+\dfrac{\pi}{4}$.

   Concept and Intuition

   Both $\tan\theta$ and $\cot\theta$ have period $\pi$, so any solution set for an equation built purely from them should also have period $\pi$ (not $2\pi$). This immediately makes options with a $2n\pi$ structure suspicious, and testing values confirms which option is right.

   Step-by-Step Solution

   1. Rewrite the RHS using $\cos x=\sin\left(\dfrac{\pi}{2}-x\right)$: the equation becomes $\sin\left(\dfrac{\pi}{4}\cot\theta\right)=\sin\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\tan\theta\right)$.
   2. A natural guess is $\cot\theta=\tan\theta$, i.e. $\tan^2\theta=1 \Rightarrow \tan\theta=\pm1$, giving $\theta=\dfrac{\pi}{4}$ as a base solution (then both sides read $\sin(\pi/4)=\sqrt2/2$).
   3. Since $\tan\theta$ and $\cot\theta$ each repeat every $\pi$, check $\theta=\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}$: $\tan\theta=1,\cot\theta=1$ again, so the equation again holds. …

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7. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $1-\cot 23^0 = \dfrac{x}{1-\cot 22^0}$, then x = (A) 1 (B) 2 (C) $\dfrac{1}{2}$ (D) 3

   ›Reveal solutionSolution

   Recognizing $23^\circ+22^\circ=45^\circ$ triggers a standard cotangent identity that instantly evaluates the product without needing actual trig values.

   Concept and Intuition

   Whenever $A+B=45^\circ$, the identity $(\cot A-1)(\cot B-1)=2$ holds — this follows directly from the cotangent addition formula $\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1$.

   Step-by-Step Solution

   1. Rearrange the given equation: $x=(1-\cot23^\circ)(1-\cot22^\circ)$.
   2. Note $23^\circ+22^\circ=45^\circ$.
   3. From $\cot(A+B)=1$ when $A+B=45^\circ$: $\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1 \Rightarrow \cot A\cot B-1=\cot A+\cot B$.
   4. So $\cot A\cot B-\cot A-\cot B+1 = (\cot A+\cot B+1)-\cot A-\cot B+1=2$, i.e. $(\cot A-1)(\cot B-1)=2$. …

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8. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} =$ (A) $2\sec x$ (B) $2\csc x$ (C) $\tan 2x$ (D) $\sin 2x$

   ›Reveal solutionSolution

   Combining the two fractions and using the Pythagorean identity collapses everything to $2\csc x$.

   Concept and Intuition

   Whenever you see a sum of a fraction and its "flipped" reciprocal-like partner, combining over a

   common denominator and using $\sin^2x+\cos^2x=1$ is almost always the fastest route — the

   $(1+\cos x)^2$ expansion conveniently reintroduces $\sin^2x+\cos^2x$, letting everything collapse.

   Step-by-Step Solution

   1. Write the sum with common denominator $\sin x(1+\cos x)$:

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

   2. Expand the numerator: $\sin^2x + 1+2\cos x+\cos^2x = (\sin^2x+\cos^2x)+1+2\cos x = 1+1+2\cos x = 2+2\cos x$. …

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9. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.The solution of the equation $[\sin x + \cos x]^{1+\sin 2x} = 2$, where $-\pi \le x \le \pi$ is (A) $\dfrac{\pi}{2}$ (B) $\pi$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{3\pi}{4}$

   ›Reveal solutionSolution

   This tests recognizing $1+\sin 2x=(\sin x+\cos x)^2$ and that $\sin x+\cos x$ has maximum value

   $\sqrt2$; the answer is $x=\dfrac{\pi}{4}$.

   Concept and Intuition

   The expression $[\sin x+\cos x]^{1+\sin 2x}=2$ looks intimidating because both the base and the

   exponent depend on $x$. The key simplification is to notice that the exponent is not independent of

   the base — it is literally the square of the base. Since

   $(\sin x+\cos x)^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin 2x,$

   the whole equation collapses to a single-variable relation once we substitute $t=\sin x+\cos x$.

   Step-by-Step Solution

   1. Let $t=\sin x+\cos x=\sqrt2\sin\!\left(x+\dfrac{\pi}{4}\right)$, so $t\in[-\sqrt2,\sqrt2]$.
   2. The exponent $1+\sin2x=t^2$, so the equation is $t^{t^2}=2$.
   3. Since $t\le \sqrt2$, try the extreme value $t=\sqrt2$: then $t^2=(\sqrt2)^2=2$, and $t^{t^2}=(\sqrt2)^2=2$. This satisfies the equation exactly.
   4. $t=\sqrt2$ means $\sqrt2\sin\!\left(x+\dfrac{\pi}{4}\right)=\sqrt2\Rightarrow \sin\!\left(x+\dfrac{\pi}{4}\right)=1$.
   5. So $x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+2k\pi\Rightarrow x=\dfrac{\pi}{4}+2k\pi$. Within $-\pi\le x\le\pi$, the only solution is $x=\dfrac{\pi}{4}$. …

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10. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.$\dfrac{1}{1+\sin\theta}+\dfrac{1}{1-\sin\theta} =$ (A) $2\cos^2\theta$ (B) $-2\cos^2\theta$ (C) $2\tan^2\theta$ (D) $2\sec^2\theta$

    ›Reveal solutionSolution

    This tests combining two fractions over a common denominator and simplifying using the Pythagorean identity $1-\sin^2\theta=\cos^2\theta$. Answer: $2\sec^2\theta$.

    Concept and Intuition

    Adding fractions with denominators that are conjugates of each other, $(1+\sin\theta)$ and $(1-\sin\theta)$, produces a difference-of-squares denominator, which simplifies beautifully using the fundamental identity $\sin^2\theta+\cos^2\theta=1$.

    Step-by-Step Solution

    1. Find a common denominator: $\dfrac{1}{1+\sin\theta}+\dfrac{1}{1-\sin\theta} = \dfrac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}$.
    2. Numerator simplifies: $(1-\sin\theta)+(1+\sin\theta) = 2$.
    3. Denominator is a difference of squares: $(1+\sin\theta)(1-\sin\theta) = 1-\sin^2\theta$.
    4. Use $1-\sin^2\theta=\cos^2\theta$: denominator becomes $\cos^2\theta$. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.The value of $\tan\left(\dfrac{7\pi}{8}\right)$ is (A) $\sqrt{2}-1$ (B) $1-\sqrt{2}$ (C) $1+\sqrt{2}$ (D) $\dfrac{1}{1+\sqrt{2}}$

    ›Reveal solutionSolution

    Using $\tan(\pi-\theta)=-\tan\theta$ and the known value $\tan\frac{\pi}{8}=\sqrt2-1$, we get $\tan\frac{7\pi}{8}=1-\sqrt2$.

    Concept and Intuition

    Angles in the second quadrant can always be reduced to a first-quadrant reference angle using supplementary-angle identities; here $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$.

    Step-by-Step Solution

    1. Write $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$.
    2. Use $\tan(\pi-\theta) = -\tan\theta$, so $\tan\frac{7\pi}{8} = -\tan\frac{\pi}{8}$.
    3. Recall (or derive from half-angle formula with $\theta=\pi/4$): $\tan\frac{\pi}{8} = \tan22.5^\circ = \sqrt2-1$. …

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12. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.$\cos\theta(\csc\theta - \sec\theta) - \cot\theta =$ (A) -1 (B) 1 (C) 0 (D) $\cos^2\theta - \tan^2\theta$

    ›Reveal solutionSolution

    Direct expansion into sines and cosines shows the $\cot\theta$ terms cancel, leaving $-1$.

    Concept and Intuition

    When an expression mixes several different trig ratios ($\csc,\sec,\cot$), it's usually fastest to

    rewrite everything in terms of $\sin\theta$ and $\cos\theta$ and simplify directly, rather than

    hunting for a named identity.

    Step-by-Step Solution

    1. Expand: $\cos\theta(\csc\theta-\sec\theta) = \cos\theta\csc\theta - \cos\theta\sec\theta$.
    2. $\cos\theta\csc\theta = \cos\theta\cdot\dfrac{1}{\sin\theta} = \dfrac{\cos\theta}{\sin\theta}=\cot\theta$.
    3. $\cos\theta\sec\theta = \cos\theta\cdot\dfrac1{\cos\theta}=1$.
    4. So the expression becomes $\cot\theta - 1 - \cot\theta = -1$.

    Common Mistakes …

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