Trigonometrical ratios of sum and difference of two and three angles Questions in English

(B) We use the sum and difference formulas for hyperbolic functions:
$\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$
$\sinh(x-y) = \sinh x \cosh y - \cosh x \sinh y$
$\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y$
$\cosh(x-y) = \cosh x \cosh y - \sinh x \sinh y$
Substituting these into the numerator:
$\sinh(x+y) + \sinh(x-y) = (\sinh x \cosh y + \cosh x \sinh y) + (\sinh x \cosh y - \cosh x \sinh y) = 2 \sinh x \cosh y$
Substituting these into the denominator:
$\cosh(x+y) - \cosh(x-y) = (\cosh x \cosh y + \sinh x \sinh y) - (\cosh x \cosh y - \sinh x \sinh y) = 2 \sinh x \sinh y$
Now,divide the numerator by the denominator:
$\frac{2 \sinh x \cosh y}{2 \sinh x \sinh y} = \frac{\cosh y}{\sinh y} = \coth y$
Therefore,the correct option is $B$.

(A) Given the equation: $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$
Using the trigonometric identity $\cos (x-y) = \cos x \cos y + \sin x \sin y$,where $x = A+B$ and $y = A-B$:
$\cos ((A+B) - (A-B)) = \frac{1}{2}$
$\cos (A+B-A+B) = \frac{1}{2}$
$\cos (2B) = \frac{1}{2}$
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have $2B = \frac{\pi}{3}$
Therefore,$B = \frac{\pi}{6}$
Thus,option $A$ is correct.

(C) Given: $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$
Rearranging the terms,we get:
$m [\cos (\alpha+\beta) - \cos (\alpha-\beta)] = n [\cos (\alpha+\beta) + \cos (\alpha-\beta)]$
Using the identities $\cos (A+B) - \cos (A-B) = -2 \sin A \sin B$ and $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$:
$m [-2 \sin \alpha \sin \beta] = n [2 \cos \alpha \cos \beta]$
$-2m \sin \alpha \sin \beta = 2n \cos \alpha \cos \beta$
Dividing both sides by $2m \cos \alpha \cos \beta$ (assuming $\cos \alpha \cos \beta \neq 0$):
$\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}$
$\tan \alpha \tan \beta = -\frac{n}{m}$

(A) Given: $\sin B = \frac{5}{13}$. Since $B$ is an acute angle,$\cos B = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$. Thus,$\tan B = \frac{5}{12}$.
Given: $\sin (A-B) = \frac{16}{65}$. Since $A$ and $B$ are acute,$\cos (A-B) = \sqrt{1 - (\frac{16}{65})^2} = \sqrt{\frac{4225-256}{4225}} = \sqrt{\frac{3969}{4225}} = \frac{63}{65}$.
Thus,$\tan (A-B) = \frac{16}{63}$.
Using the formula $\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we have:
$\frac{\tan A - \frac{5}{12}}{1 + \tan A \cdot \frac{5}{12}} = \frac{16}{63}$
$\frac{12 \tan A - 5}{12 + 5 \tan A} = \frac{16}{63}$
$63(12 \tan A - 5) = 16(12 + 5 \tan A)$
$756 \tan A - 315 = 192 + 80 \tan A$
$676 \tan A = 507$
$\tan A = \frac{507}{676} = \frac{3}{4}$.
Then $\cot A = \frac{1}{\tan A} = \frac{4}{3}$.
Finally,$\tan A + \cot A = \frac{3}{4} + \frac{4}{3} = \frac{9+16}{12} = \frac{25}{12}$.
Wait,re-evaluating the calculation: $\tan A = \frac{507}{676} = 0.75 = \frac{3}{4}$. The provided option $\frac{714025}{342732}$ is $\frac{507^2 + 676^2}{507 \times 676} = \frac{257049 + 456976}{342732} = \frac{714025}{342732}$. This matches option $A$.

(C) We know that $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Therefore,$\tan A - \tan B = \tan(A-B)(1 + \tan A \tan B)$.
Substituting $A = 80^{\circ}$ and $B = 10^{\circ}$:
$\tan 80^{\circ} - \tan 10^{\circ} = \tan(80^{\circ}-10^{\circ})(1 + \tan 80^{\circ} \tan 10^{\circ}) = \tan 70^{\circ}(1 + \tan 80^{\circ} \tan 10^{\circ})$.
Now,$\frac{\tan 80^{\circ}-\tan 10^{\circ}}{\tan 70^{\circ}} = \frac{\tan 70^{\circ}(1 + \tan 80^{\circ} \tan 10^{\circ})}{\tan 70^{\circ}} = 1 + \tan 80^{\circ} \tan 10^{\circ}$.
Since $\tan 80^{\circ} = \cot 10^{\circ}$,we have $1 + \cot 10^{\circ} \tan 10^{\circ} = 1 + 1 = 2$.

(C) Given $0 < A < B < \frac{\pi}{4}$,$\cos (A+B) = \frac{11}{61}$ and $\sin (A-B) = \frac{24}{25}$.
Since $0 < A+B < \frac{\pi}{2}$,$\sin (A+B) = \sqrt{1 - (\frac{11}{61})^2} = \sqrt{\frac{3721-121}{3721}} = \frac{60}{61}$.
Since $0 < A-B < 0$ is not possible,we note that $A < B$ implies $A-B < 0$. However,the problem states $\sin(A-B) = \frac{24}{25}$,which implies $A-B$ must be in the first or second quadrant. Given the constraints,we treat the values as magnitudes for the identity.
Using $\cos (A-B) = \sqrt{1 - (\frac{24}{25})^2} = \frac{7}{25}$.
We know $\sin 2A = \sin ((A+B) + (A-B)) = \sin (A+B) \cos (A-B) + \cos (A+B) \sin (A-B) = (\frac{60}{61} \times \frac{7}{25}) + (\frac{11}{61} \times \frac{24}{25}) = \frac{420 + 264}{1525} = \frac{684}{1525}$.
We know $\sin 2B = \sin ((A+B) - (A-B)) = \sin (A+B) \cos (A-B) - \cos (A+B) \sin (A-B) = (\frac{60}{61} \times \frac{7}{25}) - (\frac{11}{61} \times \frac{24}{25}) = \frac{420 - 264}{1525} = \frac{156}{1525}$.
Therefore,$\sin 2A + \sin 2B = \frac{684}{1525} + \frac{156}{1525} = \frac{840}{1525} = \frac{168}{305}$.

(B) Given that,$A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$.
Since $A+B+C = 35^{\circ} + 15^{\circ} + 40^{\circ} = 90^{\circ}$,we have $\tan(A+B+C) = \tan(90^{\circ})$,which is undefined.
The formula for $\tan(A+B+C)$ is given by:
$\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}$
For this to be undefined,the denominator must be zero:
$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 0$
Therefore,$\tan A \tan B + \tan B \tan C + \tan C \tan A = 1$.

(C) Given $\sin A = -\frac{24}{25}$. Since $A$ is not in the $4^{\text{th}}$ quadrant and $\sin A < 0$,$A$ must be in the $3^{\text{rd}}$ quadrant. Thus,$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (-\frac{24}{25})^2} = -\sqrt{1 - \frac{576}{625}} = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.
Given $\cos B = \frac{15}{17}$. Since $B$ is not in the $1^{\text{st}}$ quadrant and $\cos B > 0$,$B$ must be in the $4^{\text{th}}$ quadrant. Thus,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{15}{17})^2} = -\sqrt{1 - \frac{225}{289}} = -\sqrt{\frac{64}{289}} = -\frac{8}{17}$.
Now,$\sin(A+B) = \sin A \cos B + \cos A \sin B = (-\frac{24}{25})(\frac{15}{17}) + (-\frac{7}{25})(-\frac{8}{17}) = -\frac{360}{425} + \frac{56}{425} = -\frac{304}{425} < 0$.
$\cos(A+B) = \cos A \cos B - \sin A \sin B = (-\frac{7}{25})(\frac{15}{17}) - (-\frac{24}{25})(-\frac{8}{17}) = -\frac{105}{425} - \frac{192}{425} = -\frac{297}{425} < 0$.
Since both $\sin(A+B) < 0$ and $\cos(A+B) < 0$,$(A+B)$ lies in the $3^{\text{rd}}$ quadrant.

(A) We use the identity $2 \cosh A \sinh B = \sinh(A+B) + \sinh(A-B)$.
Let $A = x+y$ and $B = x-y$.
Then $A+B = (x+y) + (x-y) = 2x$ and $A-B = (x+y) - (x-y) = 2y$.
Substituting these into the expression:
$2 \cosh (x+y) \sinh (x-y) = \sinh(2x) + \sinh(2y)$.
Now,add $\sinh 2y$ to the expression:
$\sinh 2x + \sinh 2y + \sinh 2y = \sinh 2x + 2 \sinh 2y$.
Wait,re-evaluating the expression $2 \cosh A \sinh B = \sinh(A+B) - \sinh(A-B)$ is incorrect. The correct identity is $2 \cosh A \sinh B = \sinh(A+B) - \sinh(A-B)$.
Let us re-calculate:
$2 \cosh (x+y) \sinh (x-y) = \sinh((x+y) + (x-y)) - \sinh((x+y) - (x-y)) = \sinh 2x - \sinh 2y$.
Adding $\sinh 2y$ to this result:
$(\sinh 2x - \sinh 2y) + \sinh 2y = \sinh 2x$.

(A) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.
Let $A = \frac{\pi}{6} + \theta$ and $B = \frac{\pi}{6} - \theta$.
Then $A+B = \left(\frac{\pi}{6} + \theta\right) + \left(\frac{\pi}{6} - \theta\right) = \frac{2\pi}{6} = \frac{\pi}{3}$.
And $A-B = \left(\frac{\pi}{6} + \theta\right) - \left(\frac{\pi}{6} - \theta\right) = 2\theta$.
Substituting these into the identity:
$\cos^2\left(\frac{\pi}{6}+\theta\right)-\sin^2\left(\frac{\pi}{6}-\theta\right) = \cos\left(\frac{\pi}{3}\right) \cos(2\theta)$.
Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$,the expression equals $\frac{1}{2} \cos 2\theta$.

(A) Let the starting point be $O$. After $2 \text{ hours}$,the distance travelled by the first ship is $OA = 3 \text{ km/h} \times 2 \text{ h} = 6 \text{ km}$.
The distance travelled by the second ship is $OB = 4 \text{ km/h} \times 2 \text{ h} = 8 \text{ km}$.
The angle between the two paths is $\angle AOB = 45^{\circ} + 15^{\circ} = 60^{\circ}$.
Using the Law of Cosines in $\triangle AOB$:
$AB^2 = OA^2 + OB^2 - 2(OA)(OB) \cos(60^{\circ})$
$AB^2 = 6^2 + 8^2 - 2(6)(8) \times \frac{1}{2}$
$AB^2 = 36 + 64 - 48$
$AB^2 = 100 - 48 = 52$
$AB = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13} \text{ km}$.

(C) We know that $\cos 75^{\circ} = \cos(45^{\circ}+30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} - \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Similarly,$\cos 15^{\circ} = \cos(45^{\circ}-30^{\circ}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Now,$\cos^2 75^{\circ} + \cos^2 15^{\circ} = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^2 = \frac{3+1-2\sqrt{3}}{8} + \frac{3+1+2\sqrt{3}}{8} = \frac{8}{8} = 1$.
Also,$\cos^2 45^{\circ} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.
And $\cos^2 30^{\circ} + \cos^2 60^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$.
Substituting these values into the expression: $1 + \frac{1}{2} - 1 = \frac{1}{2}$.

(A) Given $\cos (\theta+\phi) = \frac{3}{5}$,since $0 < \theta, \phi < \frac{\pi}{4}$,we have $\tan (\theta+\phi) = \frac{4}{3}$.
Given $\sin (\theta-\phi) = \frac{5}{13}$,we have $\tan (\theta-\phi) = \frac{5}{12}$.
We know that $2\theta = (\theta+\phi) + (\theta-\phi)$.
Therefore,$\tan (2\theta) = \tan ((\theta+\phi) + (\theta-\phi)) = \frac{\tan (\theta+\phi) + \tan (\theta-\phi)}{1 - \tan (\theta+\phi) \tan (\theta-\phi)}$.
Substituting the values: $\tan (2\theta) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3} \times \frac{5}{12})} = \frac{\frac{16+5}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{21}{12} \times \frac{36}{16} = \frac{21 \times 3}{16} = \frac{63}{16}$.
Thus,$\cot (2\theta) = \frac{1}{\tan (2\theta)} = \frac{16}{63}$.

(A) We use the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.
Substituting the given values:
$\tan(\alpha + \beta) = \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \left(\frac{a}{a+1}\right) \left(\frac{1}{2a+1}\right)}$
$= \frac{\frac{a(2a+1) + 1(a+1)}{(a+1)(2a+1)}}{\frac{(a+1)(2a+1) - a}{(a+1)(2a+1)}}$
$= \frac{2a^2 + a + a + 1}{2a^2 + a + 2a + 1 - a}$
$= \frac{2a^2 + 2a + 1}{2a^2 + 2a + 1} = 1$
Since $\tan(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

(B) Given that $\theta + \phi = \frac{\pi}{4}$.
Taking $\tan$ on both sides,we get $\tan(\theta + \phi) = \tan(\frac{\pi}{4}) = 1$.
Using the formula $\tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} = 1$.
This implies $\tan \theta + \tan \phi = 1 - \tan \theta \tan \phi$,or $\tan \theta + \tan \phi + \tan \theta \tan \phi = 1$.
Now,consider the expression $(1 + \tan \theta)(1 + \tan \phi) = 1 + \tan \phi + \tan \theta + \tan \theta \tan \phi$.
Substituting the value from the previous step: $1 + (\tan \theta + \tan \phi + \tan \theta \tan \phi) = 1 + 1 = 2$.

(B) We know that $\cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Also,$\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.
Subtracting the two values:
$\cos 15^{\circ} - \sin 15^{\circ} = \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(C) Given $\cos(\alpha+\beta) = -\frac{1}{10}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$,we have $0 < \alpha+\beta < \frac{7\pi}{12}$. Since $\cos(\alpha+\beta) < 0$,we must have $\frac{\pi}{2} < \alpha+\beta < \frac{7\pi}{12}$.
Thus,$\sin(\alpha+\beta) = \sqrt{1 - (-\frac{1}{10})^2} = \sqrt{\frac{99}{100}} = \frac{3\sqrt{11}}{10}$.
So,$\tan(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$.
Given $\sin(\alpha-\beta) = \frac{3}{8}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$,we have $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{3}$. Since $\sin(\alpha-\beta) > 0$,we have $0 < \alpha-\beta < \frac{\pi}{3}$.
Thus,$\cos(\alpha-\beta) = \sqrt{1 - (\frac{3}{8})^2} = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8}$.
So,$\tan(\alpha-\beta) = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}}$.
Now,$\tan 2\alpha = \tan((\alpha+\beta)+(\alpha-\beta)) = \frac{\tan(\alpha+\beta) + \tan(\alpha-\beta)}{1 - \tan(\alpha+\beta)\tan(\alpha-\beta)}$.
Substituting the values,$\tan 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{55}}}{1 - (-3\sqrt{11})(\frac{3}{\sqrt{55}})} = \frac{\frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}}}{1 + \frac{9\sqrt{11}}{\sqrt{55}}} = \frac{-3(11\sqrt{5}) + 3}{\sqrt{55} + 9\sqrt{11}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(\sqrt{5} + 9)}$.
Comparing with $\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$,we get $r=11$ and $s=9$.
Therefore,$r+s = 11+9 = 20$.