Q.Find the value of $\tan\frac{13\pi}{12}$.

From $\sin\theta=\tfrac35$, $\cos^{2}\theta=1-\tfrac{9}{25}=\tfrac{16}{25}$, so $\cos\theta=\pm\tfrac45$. Given $\cos\theta<0$, $\cos\theta=-\tfrac45$.

Sign check by quadrant:

• I: $\sin>0,\cos>0\Rightarrow$ product $>0$. ✓
• II: $\sin>0,\cos<0\Rightarrow$ product $<0$.
• III: $\sin<0,\cos<0\Rightarrow$ product $>0$. ✓
• IV: $\sin<0,\cos>0\Rightarrow$ product $<0$.

So $\theta$ lies in the first or the third quadrant.

Evaluate each term using periodicity and sign rules:

• $\cot(-405^\circ)=-\cot405^\circ=-\cot45^\circ=-1$.
• $\tan315^\circ=\tan(315^\circ-360^\circ)=\tan(-45^\circ)=-1$.
• $\cot495^\circ=\cot(495^\circ-360^\circ)=\cot135^\circ=-1$.
• $\tan(-585^\circ)=-\tan585^\circ=-\tan(585^\circ-360^\circ)=-\tan225^\circ=-\tan45^\circ=-1$.

Substitute:

Concept and Intuition

On (pi/2, 3pi/2), sine is positive only in the second quadrant (pi/2, pi), where cosine is negative. That fixes the sign of the tangent.

Step-by-Step Solution

1. sin alpha = 12/13 > 0 with pi/2 < alpha < 3pi/2 forces alpha in (pi/2, pi).
2. cos alpha = -sqrt(1 - 144/169) = -5/13.
3. tan alpha = sin/cos = (12/13)/(-5/13) = -12/5.

Common Mistakes

• Taking cos = +5/13 and getting +12/5, ignoring the quadrant.

ANSWER: C

Find $\cos\theta$. $\cos^{2}\theta=1-\frac{1}{25}=\frac{24}{25}$. In the second quadrant cosine is negative, so $\cos\theta=-\frac{\sqrt{24}}{5}=-\frac{2\sqrt6}{5}$.

Secant. $\sec\theta=\frac{1}{\cos\theta}=-\frac{5}{2\sqrt6}$.

Evaluate each factor.

$\tan315^\circ=\tan(360^\circ-45^\circ)=-\tan45^\circ=-1.$

For $\cot(-405^\circ)$, add $720^\circ$ (period of cotangent is $180^\circ$, but adding full turns is safe): $-405^\circ+720^\circ=315^\circ$, so $\cot(-405^\circ)=\cot315^\circ=\dfrac{\cos315^\circ}{\sin315^\circ}=\dfrac{\tfrac{\sqrt2}{2}}{-\tfrac{\sqrt2}{2}}=-1.$

Therefore

$1-\tan x=\dfrac{2}{3+\sqrt3}=\dfrac{2(3-\sqrt3)}{6}=\dfrac{3-\sqrt3}{3}=1-\dfrac{1}{\sqrt3}$, so $\tan x=\dfrac{1}{\sqrt3}$. With $0\le x<\tfrac\pi2$, $x=\dfrac\pi6$.

Set $\theta=\cos^{-1}\left(\dfrac{-24}{25}\right)$. The range of $\cos^{-1}$ is $[0,\pi]$, and since the cosine is negative, $\theta$ is in $(\pi/2,\pi)$ — the second quadrant, where $\sin\theta>0$.

Then $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\dfrac{576}{625}}=\sqrt{\dfrac{49}{625}}=\dfrac{7}{25}$.

Therefore $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{7/25}{-24/25}=-\dfrac{7}{24}$.

From $\cos x-\sin x=0$ we get $\tan x=1$. The general solution is $x=\dfrac{\pi}{4}+n\pi$.

Within $0\leq x\leq \pi$, the candidate $\dfrac{5\pi}{4}$ exceeds $\pi$, so the only valid value is

From $2\sin\!\left(\dfrac{\pi}{3}-2x\right)-1=0$ we get $\sin\!\left(\dfrac{\pi}{3}-2x\right)=\dfrac12$.

So $\dfrac{\pi}{3}-2x=\dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$.

• $\dfrac{\pi}{3}-2x=\dfrac{\pi}{6}\Rightarrow 2x=\dfrac{\pi}{6}\Rightarrow x=\dfrac{\pi}{12}$.
• $\dfrac{\pi}{3}-2x=\dfrac{5\pi}{6}\Rightarrow 2x=-\dfrac{\pi}{2}\Rightarrow x=-\dfrac{\pi}{4}$ (rejected, not in range).

Hence $x=\dfrac{\pi}{12}$.

Given $\sin x=\dfrac35$, so $\cos x=\sqrt{1-\tfrac{9}{25}}=\dfrac45$ (taking the positive value).

Then $\sec x=\dfrac{1}{\cos x}=\dfrac54$ and $\tan x=\dfrac{\sin x}{\cos x}=\dfrac{3}{4}$.

Concept and Intuition

Use $\sin^2\theta+\cos^2\theta=1$ and fix the sign from the quadrant implied by the sign of $\tan\theta$.

Step-by-Step Solution

1. $\cos\theta=\frac{5}{11}>0$; $\tan\theta<0$ means $\sin\theta$ and $\cos\theta$ have opposite signs, so $\sin\theta<0$ (quadrant IV).
2. $\sin^2\theta=1-\frac{25}{121}=\frac{96}{121}$.
3. $\sin\theta=-\sqrt{\frac{96}{121}}=-\frac{\sqrt{96}}{11}$.
4. $\sqrt{96}=\sqrt{16\cdot6}=4\sqrt6$, so $\sin\theta=-\frac{4\sqrt6}{11}$.

Common Mistakes

• Taking the positive root and ignoring the quadrant.
• Not simplifying $\sqrt{96}=4\sqrt6$.

ANSWER: D