Let C be the position of the target and A and B be the positions of the fighter jet

Given ㄥBAC = 30, ㄥABC = 45
ஃ ㄥC = 180 - (30 - 45) = 180 - 75 = 105
Given AB = 100 km
Using sine formula,
\(\frac { a }{ sinA } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin30° } =\frac { 100 }{ sin105° } \)
\(\Rightarrow \frac { a }{ \frac { 1 }{ 2 } } =\frac { 100 }{ sin105° } \Rightarrow 2a=\frac { 100 }{ sin105° } \Rightarrow a=\frac { 50 }{ sin105° } \)
Now, sin 105° = sin (60 + 45) = sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)
Substituting (2) in (1) we get,
a = \(\frac { 50 }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } \Rightarrow a=\frac { 50\left( 2\sqrt { 2 } \right) }{ \sqrt { 3 } +1 } =\frac { 100\sqrt { 2 } }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
a = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

Given AC = 4 km, ㄥA = 60°

and ㄥB = 45
ㄥC = 180 - (A + B)
= 180 - (60 + 45)
= 180 - 105 = 75
Using sine formula,
\(\frac { a }{ sinA } =\frac { b }{ sinB } \)
\(\frac { a }{ \frac { \sqrt { 3 } }{ 2 } } =\frac { 4 }{ \frac { 1 }{ \sqrt { 2 } } } \)
\(\Rightarrow \frac { a }{ sin60° } =\frac { 4 }{ sin45° } \)
\(\Rightarrow \frac { 2a }{ \sqrt { 3 } } =4\sqrt { 2 } \)
\(\Rightarrow a=\frac { 4\sqrt { 6 } }{ 2 } =2\sqrt { 6 } \)
Again using sine formula,
\(\frac { c }{ sinC } =\frac { a }{ sinA } \)
\(\Rightarrow \frac { c }{ sin75° } =\frac { 2\sqrt { 6 } }{ sin60° } \)
\(\Rightarrow \frac { c }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } =\frac { 2\sqrt { 6 } }{ \frac { \sqrt { 3 } }{ 2 } } \)
\(\Rightarrow \frac { C.2\sqrt { 2 } }{ \sqrt { 3 } +1 } =\frac { 4\sqrt { 6 } }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 4\sqrt { 6 } \left( \sqrt { 3 } +1 \right) }{ 2\sqrt { 6 } } =2\left( \sqrt { 3 } +1 \right) \)
ஃ Total distance covered by the man
= a + b + c
= 2√6 + 4 + 2(√3 + 1)
= 2√6 + 4 + 2√3 + 2
= (6 + 2√3 + 2√6)km

Given \(\frac { cos^{ 4 }\alpha }{ { cos }^{ 2 }\beta } +\frac { { sin }^{ 4 }\alpha }{ { sin }^{ 2 }\beta } \)
⇒ cos⁴\(\alpha\)sin²β + sin⁴\(\alpha\)cos²β = cos²βsin²β
⇒ cos⁴\(\alpha\)(1-cos²β) + cos²β(1-cos²\(\alpha\))2 = cos²β(1-cos²β)

⇒ cos⁴\(\alpha\)-2cos²\(\alpha\)cos²β+cos⁴β = 0
⇒ (cos²\(\alpha\)-cos²β)² = 0
⇒ cos²\(\alpha\)-cos²β = 0
⇒ cos²\(\alpha\) = cos²β...(1)
⇒ 1-sin²\(\alpha\) = 1-sin²β
⇒ sin²\(\alpha\) = sin^-2β
\(\frac { cos^{ 4 }\alpha }{ { cos }^{ 2 }\beta } +\frac { { sin }^{ 4 }\alpha }{ { sin }^{ 2 }\beta } =1\)
LHS = \(\frac { { cos }^{ 4 }\beta }{ { cos }^{ 2 }\alpha } +\frac { { sin }^{ 4 }\beta }{ { sin }^{ 2 }\alpha } \)
= \(\frac { { cos }^{ 2 }\beta { cos }^{ 2 }\beta }{ { cos }^{ 2 }\alpha } +\frac { { sin }^{ 2 }\beta { sin }^{ 2 }\beta }{ { sin }^{ 2 }\alpha } \)
\(={ cos }^{ 2 }\beta +{ sin }^{ 2 }\beta =1\) = RHS

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.

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

LHS = cos A + cos B - cos C
\(=2cos\left( cos\frac { A+B }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) -\left( 1-2{ sin }^{ 2 }\frac { c }{ 2 } \right) \)
\(A+B+C=180\Rightarrow A+B=180-C\Rightarrow \frac { A+B }{ 2 } =90-\frac { C }{ 2 } \)
\(\therefore cos\left( \frac { A+B }{ 2 } \right) =cos\left( 90-\frac { C }{ 2 } \right) =sin\frac { C }{ 2 } \)
\(\therefore LHS=2sin\frac { C }{ 2 } cos\left( \frac { A-B }{ 2 } \right) -1+2sin\frac { C }{ 2 } \)
\(=2sin\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +sin\frac { C }{ 2 } \right] -1\)
\(=-1+2sin\frac { C }{ 2 } \left[ \left( cos\frac { A-B }{ 2 } \right) cos\left( \frac { A+B }{ 2 } \right) \right] \)
\(=-1+2sin\frac { C }{ 2 } \left[ 2cos\frac { A }{ 2 } cos\frac { B }{ 2 } \right] \)
\(=-1+4cos\frac { A }{ 2 } cos\frac { B }{ 2 } sin\frac { C }{ 2 } \)
= RHS
= Hence Proved

Let x cosθ = \(y\quad cos\left( θ+2\frac { \pi }{ 3 } \right) =z\quad cos\left( θ+2\frac { \pi }{ 3 } \right) =\lambda \)
\(\frac { \lambda }{ x } =cos\theta ,\frac { \lambda }{ y } =cos\left( θ+2\frac { \pi }{ 3 } \right) \quad \)and \(\frac { \lambda }{ z } =cos\left( θ+4\frac { \pi }{ 3 } \right) \)
\(\frac { \lambda }{ x } +\frac { \lambda }{ y } +\frac { \lambda }{ z } =cos\theta +cos\left( θ+2\frac { \pi }{ 3 } \right) +cos\left( θ+4\frac { \pi }{ 3 } \right) \)
= cos θ + cos(120 + θ) + cos(240 + θ)
= cos θ + cos120 cos θ - sin 120 sin θ + cos 240 cos θ - sin 240 sin θ
= cos θ - cos 60 cos θ - sin 60 sin θ - sin 30 cos θ + cos 30 sin θ
= cos θ - \(\frac{1}{2}cosθ-\frac{\sqrt{3}}{2}sinθ-\frac{1}{2}cosθ+\frac{\sqrt{3}}{2}sinθ\)
= cos θ - \(\frac{1}{2}cosθ-\frac{1}{2}cosθ\)
= cos θ - cos θ = 0
\(\therefore \ \lambda =\left( \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \right) =0\Rightarrow \lambda \left( \frac { yz+xz+xy }{ xyz } \right) =0\)
⇒ xy + yz + zx = 0

LHS = sin A + sin B + sin C
= \(2sin\left( \frac { A+B }{ 2 } \right) .cos\left( \frac { A-B }{ 2 } \right) +2sin\frac { C }{ 2 } cos\frac { C }{ 2 } \)
= \(2sin\left( 90-\frac { C }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) +2sin\frac { C }{ 2 } cos\frac { C }{ 2 } \)
= \(2cos\frac { C }{ 2 } cos\left( \frac { A-B }{ 2 } \right) +2sin\frac { C }{ 2 } cos\frac { C }{ 2 } \)
= \(2cos\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +sin\frac { C }{ 2 } \right] \)
= \(2cos\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +sin\left( 90-\left( \frac { A+B }{ 2 } \right) \right) \right] \)
= \(2cos\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +cos\left( \frac { A+B }{ 2 } \right) \right] \)
= \(2cos\frac { C }{ 2 } .cos\frac { A }{ 2 } cos\frac { B }{ 2 } \)
= \(4cos\frac { A }{ 2 } cos\frac { B }{ 2 } cos\frac { C }{ 2 } \) = RHS
Hence proved.

Let x = tan A, y = tan B, z = tan C
Then x + y + z - xyz = 0
tan A + tan B + tan C - tan A tan B tan C = 0
tan(A + B + C) = 0
tan2(A + B + C) = 0
tan(2A + 2B + 2C) = 0
tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C...(1)
since x = tan A, tan 2A = \(\frac { 2tanA }{ 1-{ tan }^{ 2 }A } =\frac { 2x }{ 1-{ x }^{ 2 } } \)
\(\frac { 2x }{ 1-{ x }^{ 2 } } +\frac { 2y }{ 1-{ y }^{ 2 } } +\frac { 2z }{ 1-{ z }^{ 2 } } =\frac { 2x }{ 1-{ x }^{ 2 } } +\frac { 2y }{ 1-{ y }^{ 2 } } +\frac { 2z }{ 1-{ z }^{ 2 } } \)
Hence proved.