Given sec θ + tan θ = p ...(1)
We know sec²θ-tan²θ = 1
(sec θ + tan θ) (sec θ - tan θ) = 1
p(sec θ - tan θ) = 1
sec θ - tan θ = \(\frac{1}{p}\)...(2)
(1)+(2)➝ (sec θ + tan θ) + (sec θ - tan θ) = p+\(\frac{1}{p}\)
2sec θ = \(\frac{p^2+1}{2p}\)...(3)
(1)-(2)⟶ (sec θ + tan θ) - (sec θ - tan θ) = p-\(\frac{1}{p}\)
2tan θ = \(\frac{p^2-1}{2p}\)
tan θ = \(\frac{p^2-1}{2p}\)...(4)
(4)+(3) gives,
\(\frac{tan\theta}{sec\theta}=\frac{p^2-1}{2p}\div\frac{p^2+1}{2p}\)
\(\frac{sin\theta}{cos\theta.\frac{1}{cos\theta}}=\frac{p^2-1}{2p}\times\frac{2p}{p^2+1}=\frac{p^2-1}{p^2+1}\)
Sin \(\theta\) = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

a sec θ - c tan θ = b...(1)
b sec θ + d tan θ = c...(2)

tan θ = \(\frac{ac-b^2}{bc+ad}\)

secθ = \(\frac{bd+c^2}{ad+bc}\)
We know sec²θ - tan²θ = 1
⇒ \((\frac{bd+c^2}{ad+bc})^2-(\frac{ac-b^2}{bc+ad})^2=1\)
⇒ \(\frac{(bd+c^2)^2}{(ad+bc)^2}-\frac{(ac-b^2)^2}{(bc+ad)^2}=1\)
⇒ \(\frac{(bd+c^2)^2-(ac-b^2)^2}{(ad+bc)^2}=1\)
⇒ (bd + c²)²- (ac - b²)² = (ad + bc)²
⇒ (c²+bd)² = (ad + bc)² + (ac - b²)²
Thus θ is eliminated.

\(LHS=cot(A+{ 15 }^{ 0 })-tan(A-{ 15 }^{ 0 })\)
\(=\frac { cos(A+15) }{ sin(A+15) } -\frac { sin(A-15) }{ cos(A-15) } \)
\(=\frac { cos(A+15)cos(A-15)-sin(A-15)sin(A+15) }{ sin(A+15).cos(A-15) } \)
\(=\frac { { cos }^{ 2 }A-{ sin }^{ 2 }15\left[ { sin }^{ 2 }A-{ sin }^{ 2 }{ 15 }^{ 0 } \right] }{ \frac { 1 }{ 2 } \left[ sin(A+15+A-15)+sin(A+15-A+15) \right] } \)
\(\left[ \because cos(A+B)cos(A-B)={ cos }^{ 2 }A-{ sin }^{ 2 }Bsin(A+B)sin(A-B)={ sin }^{ 2 }A-{ sin }^{ 2 }B\quad and sinAcosB=\frac { 1 }{ 2 } (sin(A+B)+sin(A-B)) \right] \)
\(=\frac { 2\left[ { cos }^{ 2 }A-{ sin }^{ 2 }15-{ sin }^{ 2 }A+{ sin }^{ 2 }{ 15 }^{ 0 } \right] }{ sin(2A)+sin({ 30 }^{ 0 }) } \quad \left[ \because cos2A={ cos }^{ 2 }A-{ sin }^{ 2 }B \right] \)
\(=\frac { 2\left( { cos }^{ 2 }A-{ sin }^{ 2 }A \right) }{ sin2A+\frac { 1 }{ 2 } } =\frac { 2,cos2A\times 2 }{ 2sin2A+1 } \)
\(=\frac { 4cos2A }{ 1+2sin2A } =RHS\)

Given\(A+B+C={ 180 }^{ 0 }\)
\(A+B={ 180 }-C\Rightarrow \frac { A }{ 2 } +\frac { B }{ 2 } =90-\frac { C }{ 2 } \)
\(tan\left( \frac { A }{ 2 } +\frac { B }{ 2 } \right) =tan\left( 90-\frac { C }{ 2 } \right) =cot\frac { C }{ 2 } \)
\(\Rightarrow \frac { tan\frac { A }{ 2 } +tan\frac { B }{ 2 } }{ 1-tan\frac { A }{ 2 } .tan\frac { B }{ 2 } } =\frac { 1 }{ tan\frac { C }{ 2 } } \)
\(tan\frac { A }{ 2 } tan\frac { C }{ 2 } +tan\frac { B }{ 2 } tan\frac { C }{ 2 } =1-tan\frac { A }{ 2 } tan\frac { B }{ 2 } \)
\(tan\frac { A }{ 2 } tan\frac { C }{ 2 } +tan\frac { B }{ 2 } tan\frac { C }{ 2 } +tan\frac { A }{ 2 } tan\frac { B }{ 2 } =1\)

sin θ + sin 5θ + sin 3θ = 0
\(2sin\left( \frac { \theta +5\theta }{ 2 } \right) .cos\left( \frac { 5\theta -\theta }{ 2 } \right) +sin3\theta =0\)
2sin 3θ + cos 2θ + sin 3θ = 0
sin 3θ(2cos 2θ + 1) = 0
sin 3θ = 0 or cos 2θ = -\(\frac{1}{2}\)
case (i):
sin3θ = 0
3θ = nπ, n∈z
θ = n\(\frac{\pi}{3}\) , n∈z
case (ii):
cos 2θ = \(-\frac{1}{2}\) = -cos(\(\frac{\pi}{3}\))
= cos(π-\(\frac{\pi}{3}\))
cos 2θ = cos(\(\frac{2\pi}{3}\))
2θ = 2nπ土 2\(\frac{\pi}{3}\), n∈z
θ = nπ士\(\frac{\pi}{3}\),n∈z
Hence the solutions are \(\frac{n\pi}{3}\)or nπ士\(\frac{\pi}{3}\), n∈z

Let P be the intruder, A and B are the soldiers.
Let x be the distance between the intruder and soldier B.
In ΔABP

Given ㄥPAB
and ㄥPBC
In ΔABP, ㄥAPB
In ΔABP, using sine formula,
\(⇒\ {5\over sin\ 15^0}={x\over sin30^0}\)
\(⇒\ x{5\over sin15^0}sin30^0=5\times {1\over 2 sin15^0}\)
Now, sin 15 = sin (45 - 30)
= sin 45 cos 30 - cos 45 sin 30
\(={1\over \sqrt2}\times{\sqrt3\over 2}-{1\over \sqrt2}={\sqrt3-1\over 2\sqrt2}\)
Substituting this value in (1) we get,
\(x={5\over {2(\sqrt3-1)\over 2\sqrt2}}={5\sqrt2\over \sqrt3-1}\)

Let c be the position of the boat and A and B are the positions of the pilot.
Using cosine formula, c² =  a² + b² - 2ab cos C

⇒ 10² = a² + 6² - 2a (6) cos 60°
⇒ 100² = a² + 36² -12a\(\left(1\over 2\right)\)
⇒ a² + 36 - 6a - 100 = 0
⇒ a² - 6a - 64 = 0
\(⇒\ a^2={6\pm\sqrt{(-6)^2-4(2)(-64)}\over 2}\)  \(\begin{bmatrix} ∵\ x={-b\pm\sqrt{b^2-4ac}\over 2a} \\ a=1,b=-6,c=-64 \end{bmatrix}\)
\(⇒\ a={6\pm\sqrt{36+256}\over 2}⇒a={6\pm\sqrt{292}\over 2}\)
\(⇒\ a=3\pm\sqrt3⇒a=(3+\sqrt{73})km\)

\(y=\frac{2\ sin\alpha}{1+cos\alpha+sin\alpha}\)
\(=\frac{2sin\alpha[1+sin\alpha-cos\alpha]}{[(1+sin\alpha)+cos\alpha](1+sin\alpha-cos\alpha)}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{(1+sin\alpha)^2-cos^2\alpha}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{1+2sin\alpha+sin^2\alpha-(1-sin^2\alpha)}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{2sin\alpha+2sin^2\alpha}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{2sin\alpha(1+sin\alpha)}\)
\(=\frac{1-cos\alpha+sin\alpha}{1+sin\alpha}\)

Let AB be the chord and O be the centre of the circular park.
Let\(\angle\)AOB = \(\theta\)
Area of the segment = Area of the sector - Area of \(\triangle\)OAB

\(={1\over2}r^2\theta-{1\over2}r^2sin \theta\)
\(=({1\over 2}\times 4^2)[\theta -sin \theta] \) \(=8[\theta -sin \theta]...(i) \)
But cos \(\theta ={4^2+4^2-4^2\over 2(4)(4)}={1\over2}\)
Thus, \(\theta ={\pi\over3}\)
From (i), area of the segment to be allotted for the veterinary hospital
\(=8[{\pi\over3}-{\sqrt{3}\over2}]={4\over3}[2\pi-3\sqrt{3}]m^2\)

Let θ be the position of the cell phone from north to east of the first tower.
Then, using the cosine formula, we have \((\sqrt 31)^2=5^2+6^2-2\times5\times6cos \theta\)
31 = 25 + 36 - 60 cos θ
\(cos\theta=\frac{1}{2}\Rightarrow\theta=60^0\)
Let x be the distance of the cell phone's position from the highway.
Then, \(sin\theta=\frac{x}{5}\Rightarrow x=5\ sin\theta=5\ sin\ 60^0=\frac{5\times\sqrt 3}{2}km.\)